Class 9 Chapter Basic Concepts of Coordinate Geometry MCQs

in this blogpost we will be discuss Class 9 Chapter Basic Concepts of Coordinate Geometry (MCQs)

What is the coordinate of the origin?
- (a) (0,0)(0, 0)(0,0)
- (b) (1,1)(1, 1)(1,1)
- (c) (0,1)(0, 1)(0,1)
- (d) (1,0)(1, 0)(1,0)
Answer: (a) (0,0)(0, 0)(0,0)
Explanation: The origin in coordinate geometry is the point where both the x-axis and y-axis intersect, which is at (0,0)(0, 0)(0,0).
The point (3,−2)(3, -2)(3,−2) lies in which quadrant?
- (a) First quadrant
- (b) Second quadrant
- (c) Third quadrant
- (d) Fourth quadrant
Answer: (d) Fourth quadrant
Explanation: In the fourth quadrant, x is positive, and y is negative, which fits (3,−2)(3, -2)(3,−2).
Which axis represents the x-coordinate?
- (a) Horizontal axis
- (b) Vertical axis
- (c) Both axes
- (d) None of the above
Answer: (a) Horizontal axis
Explanation: The x-axis is horizontal, and the y-axis is vertical.
What is the distance between the points A(3,4)A(3, 4)A(3,4) and B(6,8)B(6, 8)B(6,8)?
- (a) 555
- (b) 25\sqrt{25}25
- (c) 20\sqrt{20}20
- (d) 222
Answer: (a) 555
Explanation:
Distance formula:
(x2−x1)2+(y2−y1)2\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}(x2−x1)2+(y2−y1)2
=(6−3)2+(8−4)2=32+42=9+16=25=5= \sqrt{(6 – 3)^2 + (8 – 4)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5=(6−3)2+(8−4)2=32+42=9+16=25=5.
What is the midpoint of the points A(2,5)A(2, 5)A(2,5) and B(6,3)B(6, 3)B(6,3)?
- (a) (4,4)(4, 4)(4,4)
- (b) (3,4)(3, 4)(3,4)
- (c) (5,4)(5, 4)(5,4)
- (d) (4,5)(4, 5)(4,5)
Answer: (a) (4,4)(4, 4)(4,4)
Explanation:
Midpoint formula:
(x1+x22,y1+y22)\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)(2×1+x2,2y1+y2)
=(2+62,5+32)=(4,4)= \left( \frac{2 + 6}{2}, \frac{5 + 3}{2} \right) = (4, 4)=(22+6,25+3)=(4,4).

Section 2: Distance Formula

Find the distance between A(−2,3)A(-2, 3)A(−2,3) and B(4,−1)B(4, -1)B(4,−1).Answer:
Distance formula:
(x2−x1)2+(y2−y1)2\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}(x2−x1)2+(y2−y1)2
=(4−(−2))2+(−1−3)2=(6)2+(−4)2=36+16=52=213= \sqrt{(4 – (-2))^2 + (-1 – 3)^2} = \sqrt{(6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}=(4−(−2))2+(−1−3)2=(6)2+(−4)2=36+16=52=213.
Find the distance between the points A(1,−4)A(1, -4)A(1,−4) and B(3,2)B(3, 2)B(3,2).Answer:
Distance formula:
(x2−x1)2+(y2−y1)2\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}(x2−x1)2+(y2−y1)2
=(3−1)2+(2−(−4))2=(2)2+(6)2=4+36=40=210= \sqrt{(3 – 1)^2 + (2 – (-4))^2} = \sqrt{(2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}=(3−1)2+(2−(−4))2=(2)2+(6)2=4+36=40=210.
What is the distance between the points A(0,0)A(0, 0)A(0,0) and B(7,24)B(7, 24)B(7,24)?Answer:
Distance formula:
(x2−x1)2+(y2−y1)2\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}(x2−x1)2+(y2−y1)2
=(7−0)2+(24−0)2=49+576=625=25= \sqrt{(7 – 0)^2 + (24 – 0)^2} = \sqrt{49 + 576} = \sqrt{625} = 25=(7−0)2+(24−0)2=49+576=625=25.

Section 3: Midpoint Formula

Find the midpoint of the points A(1,1)A(1, 1)A(1,1) and B(5,5)B(5, 5)B(5,5).Answer:
Midpoint formula:
(x1+x22,y1+y22)\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)(2×1+x2,2y1+y2)
=(1+52,1+52)=(3,3)= \left( \frac{1 + 5}{2}, \frac{1 + 5}{2} \right) = (3, 3)=(21+5,21+5)=(3,3).
What is the midpoint of A(−1,4)A(-1, 4)A(−1,4) and B(3,−2)B(3, -2)B(3,−2)?Answer:
Midpoint formula:
(−1+32,4+(−2)2)=(1,1)\left( \frac{-1 + 3}{2}, \frac{4 + (-2)}{2} \right) = (1, 1)(2−1+3,24+(−2))=(1,1).

Section 4: Area of Triangle in Coordinate Geometry

Find the area of the triangle formed by the points A(0,0),B(4,0),C(2,6)A(0, 0), B(4, 0), C(2, 6)A(0,0),B(4,0),C(2,6).Answer:
Area formula for triangle with coordinates A(x1,y1),B(x2,y2),C(x3,y3)A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)A(x1,y1),B(x2,y2),C(x3,y3):
Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\text{Area} = \frac{1}{2} | x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) |Area=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣
Substituting A(0,0),B(4,0),C(2,6)A(0, 0), B(4, 0), C(2, 6)A(0,0),B(4,0),C(2,6):
Area=12∣0(0−6)+4(6−0)+2(0−0)∣=12∣0+24+0∣=12×24=12\text{Area} = \frac{1}{2} | 0(0 – 6) + 4(6 – 0) + 2(0 – 0) | = \frac{1}{2} | 0 + 24 + 0 | = \frac{1}{2} \times 24 = 12Area=21∣0(0−6)+4(6−0)+2(0−0)∣=21∣0+24+0∣=21×24=12.
So, the area is 12.
Find the area of the triangle with vertices A(1,2),B(4,6),C(7,3)A(1, 2), B(4, 6), C(7, 3)A(1,2),B(4,6),C(7,3).Answer:
Using the area formula:
Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\text{Area} = \frac{1}{2} | x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) |Area=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣
Substituting A(1,2),B(4,6),C(7,3)A(1, 2), B(4, 6), C(7, 3)A(1,2),B(4,6),C(7,3):
Area=12∣1(6−3)+4(3−2)+7(2−6)∣=12∣1×3+4×1+7×(−4)∣\text{Area} = \frac{1}{2} | 1(6 – 3) + 4(3 – 2) + 7(2 – 6) | = \frac{1}{2} | 1 \times 3 + 4 \times 1 + 7 \times (-4) |Area=21∣1(6−3)+4(3−2)+7(2−6)∣=21∣1×3+4×1+7×(−4)∣
=12∣3+4−28∣=12×21=10.5= \frac{1}{2} | 3 + 4 – 28 | = \frac{1}{2} \times 21 = 10.5=21∣3+4−28∣=21×21=10.5.
So, the area is 10.5.

Section 5: Section Formula

Find the coordinates of the point dividing the line segment joining A(3,4)A(3, 4)A(3,4) and B(9,10)B(9, 10)B(9,10) in the ratio 2:3.Answer:
Section formula:
(mx2+nx1m+n,my2+ny1m+n)\left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right)(m+nmx2+nx1,m+nmy2+ny1)
=(2(9)+3(3)2+3,2(10)+3(4)2+3)= \left( \frac{2(9) + 3(3)}{2 + 3}, \frac{2(10) + 3(4)}{2 + 3} \right)=(2+32(9)+3(3),2+32(10)+3(4))
=(18+95,20+125)=(275,325)= \left( \frac{18 + 9}{5}, \frac{20 + 12}{5} \right) = \left( \frac{27}{5}, \frac{32}{5} \right)=(518+9,520+12)=(527,532).
So, the point is (275,325)\left( \frac{27}{5}, \frac{32}{5} \right)(527,532).
Find the coordinates of the point dividing the line segment joining A(1,2)A(1, 2)A(1,2) and B(5,6)B(5, 6)B(5,6) in the ratio 3:1.Answer:
Section formula:
=(3(5)+1(1)3+1,3(6)+1(2)3+1)= \left( \frac{3(5) + 1(1)}{3 + 1}, \frac{3(6) + 1(2)}{3 + 1} \right)=(3+13(5)+1(1),3+13(6)+1(2))
=(15+14,18+24)=(164,204)= \left( \frac{15 + 1}{4}, \frac{18 + 2}{4} \right) = \left( \frac{16}{4}, \frac{20}{4} \right)=(415+1,418+2)=(416,420).
So, the point is (4,5)(4, 5)(4,5).

Section 6: Application of Coordinate Geometry

If the point P(x,y)P(x, y)P(x,y) lies on the x-axis, then what is the value of yyy?Answer:
If the point lies on the x-axis, then y=0y = 0y=0.
If the point P(x,y)P(x, y)P(x,y) lies on the y-axis, then what is the value of xxx?Answer:
If the point lies on the y-axis, then x=0x = 0x=0.

Section 7: Coordinates of Points in the Plane

Find the coordinates of the midpoint of the segment joining A(−3,4)A(-3, 4)A(−3,4) and B(5,−2)B(5, -2)B(5,−2).

Answer:
Midpoint formula:
(x1+x22,y1+y22)\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)(2×1+x2,2y1+y2)
Substituting the coordinates of A(−3,4)A(-3, 4)A(−3,4) and B(5,−2)B(5, -2)B(5,−2):
=(−3+52,4+(−2)2)=(22,22)=(1,1)= \left( \frac{-3 + 5}{2}, \frac{4 + (-2)}{2} \right) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1)=(2−3+5,24+(−2))=(22,22)=(1,1).

Find the coordinates of the point dividing the line segment joining A(2,−1)A(2, -1)A(2,−1) and B(8,3)B(8, 3)B(8,3) in the ratio 4:1.

Answer:
Section formula:
(4(8)+1(2)4+1,4(3)+1(−1)4+1)\left( \frac{4(8) + 1(2)}{4 + 1}, \frac{4(3) + 1(-1)}{4 + 1} \right)(4+14(8)+1(2),4+14(3)+1(−1))
=(32+25,12−15)=(345,115)= \left( \frac{32 + 2}{5}, \frac{12 – 1}{5} \right) = \left( \frac{34}{5}, \frac{11}{5} \right)=(532+2,512−1)=(534,511).
So, the point is (345,115)\left( \frac{34}{5}, \frac{11}{5} \right)(534,511).

What are the coordinates of the centroid of the triangle formed by A(1,2),B(3,4),C(5,6)A(1, 2), B(3, 4), C(5, 6)A(1,2),B(3,4),C(5,6)?

Answer:
Centroid formula:
(x1+x2+x33,y1+y2+y33)\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)(3×1+x2+x3,3y1+y2+y3)
Substituting the coordinates of the points:
(1+3+53,2+4+63)=(93,123)=(3,4)\left( \frac{1 + 3 + 5}{3}, \frac{2 + 4 + 6}{3} \right) = \left( \frac{9}{3}, \frac{12}{3} \right) = (3, 4)(31+3+5,32+4+6)=(39,312)=(3,4).

Find the distance between A(2,3)A(2, 3)A(2,3) and B(2,7)B(2, 7)B(2,7).

Answer:
Distance formula:
(x2−x1)2+(y2−y1)2\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}(x2−x1)2+(y2−y1)2
=(2−2)2+(7−3)2=0+16=16=4= \sqrt{(2 – 2)^2 + (7 – 3)^2} = \sqrt{0 + 16} = \sqrt{16} = 4=(2−2)2+(7−3)2=0+16=16=4.

Section 8: Equations of Lines

Find the slope of the line joining the points A(1,2)A(1, 2)A(1,2) and B(3,6)B(3, 6)B(3,6).

Answer:
Slope formula:
m=y2−y1x2−x1m = \frac{y_2 – y_1}{x_2 – x_1}m=x2−x1y2−y1
=6−23−1=42=2= \frac{6 – 2}{3 – 1} = \frac{4}{2} = 2=3−16−2=24=2.
So, the slope is 2.

Find the equation of the line passing through A(1,3)A(1, 3)A(1,3) with a slope of 4.

Answer:
Using the point-slope form:
y−y1=m(x−x1)y – y_1 = m(x – x_1)y−y1=m(x−x1)
y−3=4(x−1)y – 3 = 4(x – 1)y−3=4(x−1)
y−3=4x−4y – 3 = 4x – 4y−3=4x−4
y=4x−1y = 4x – 1y=4x−1.
So, the equation is y=4x−1y = 4x – 1y=4x−1.

Find the slope of the line joining the points A(1,−1)A(1, -1)A(1,−1) and B(−3,5)B(-3, 5)B(−3,5).

Answer:
Slope formula:
m=y2−y1x2−x1m = \frac{y_2 – y_1}{x_2 – x_1}m=x2−x1y2−y1
=5−(−1)−3−1=6−4=−32= \frac{5 – (-1)}{-3 – 1} = \frac{6}{-4} = -\frac{3}{2}=−3−15−(−1)=−46=−23.
So, the slope is −32-\frac{3}{2}−23.

Find the equation of the line through the point P(2,−3)P(2, -3)P(2,−3) and parallel to the line y=3x+1y = 3x + 1y=3x+1.

Answer:
The slope of the given line is 3 (from the equation y=mx+cy = mx + cy=mx+c).
Since the line is parallel, it will have the same slope.
Using the point-slope form:
y−(−3)=3(x−2)y – (-3) = 3(x – 2)y−(−3)=3(x−2)
y+3=3x−6y + 3 = 3x – 6y+3=3x−6
y=3x−9y = 3x – 9y=3x−9.
So, the equation is y=3x−9y = 3x – 9y=3x−9.

Find the equation of the line through A(4,2)A(4, 2)A(4,2) and perpendicular to the line 2x−3y=42x – 3y = 42x−3y=4.

Answer:
The slope of the given line is found by writing the equation in slope-intercept form:
2x−3y=4⇒y=23x−432x – 3y = 4 \Rightarrow y = \frac{2}{3}x – \frac{4}{3}2x−3y=4⇒y=32x−34.
The slope of the given line is 23\frac{2}{3}32, and the slope of the line perpendicular to it will be the negative reciprocal:
m=−32m = -\frac{3}{2}m=−23.
Using the point-slope form:
y−2=−32(x−4)y – 2 = -\frac{3}{2}(x – 4)y−2=−23(x−4)
y−2=−32x+6y – 2 = -\frac{3}{2}x + 6y−2=−23x+6
y=−32x+8y = -\frac{3}{2}x + 8y=−23x+8.
So, the equation is y=−32x+8y = -\frac{3}{2}x + 8y=−23x+8.

Section 9: Slope-Intercept Form and Other Forms of the Equation of a Line

Find the equation of the line with slope 5 passing through A(1,2)A(1, 2)A(1,2).

Answer:
Using the point-slope form:
y−y1=m(x−x1)y – y_1 = m(x – x_1)y−y1=m(x−x1)
y−2=5(x−1)y – 2 = 5(x – 1)y−2=5(x−1)
y−2=5x−5y – 2 = 5x – 5y−2=5x−5
y=5x−3y = 5x – 3y=5x−3.
So, the equation is y=5x−3y = 5x – 3y=5x−3.

Find the equation of the line passing through A(0,4)A(0, 4)A(0,4) and B(2,10)B(2, 10)B(2,10).

Answer:
First, find the slope:
m=10−42−0=62=3m = \frac{10 – 4}{2 – 0} = \frac{6}{2} = 3m=2−010−4=26=3.
Using the point-slope form:
y−4=3(x−0)y – 4 = 3(x – 0)y−4=3(x−0)
y−4=3xy – 4 = 3xy−4=3x
y=3x+4y = 3x + 4y=3x+4.
So, the equation is y=3x+4y = 3x + 4y=3x+4.

Find the equation of the line passing through the point P(2,3)P(2, 3)P(2,3) and having slope 0.

Answer:
A line with slope 0 is horizontal.
The equation of the line is y=3y = 3y=3, as the point P(2,3)P(2, 3)P(2,3) lies on this line.

Find the equation of the line passing through A(1,2)A(1, 2)A(1,2) and B(4,5)B(4, 5)B(4,5).

Answer:
First, find the slope:
m=5−24−1=33=1m = \frac{5 – 2}{4 – 1} = \frac{3}{3} = 1m=4−15−2=33=1.
Using the point-slope form:
y−2=1(x−1)y – 2 = 1(x – 1)y−2=1(x−1)
y−2=x−1y – 2 = x – 1y−2=x−1
y=x+1y = x + 1y=x+1.
So, the equation is y=x+1y = x + 1y=x+1.

Find the equation of the line through P(3,4)P(3, 4)P(3,4) and perpendicular to the line y=−2x+3y = -2x + 3y=−2x+3.

Answer:
The slope of the given line is −2-2−2, and the slope of the perpendicular line will be the negative reciprocal:
m=12m = \frac{1}{2}m=21.
Using the point-slope form:
y−4=12(x−3)y – 4 = \frac{1}{2}(x – 3)y−4=21(x−3)
y−4=12x−32y – 4 = \frac{1}{2}x – \frac{3}{2}y−4=21x−23
y=12x+52y = \frac{1}{2}x + \frac{5}{2}y=21x+25.
So, the equation is y=12x+52y = \frac{1}{2}x + \frac{5}{2}y=21x+25.

Section 10: Complex Problems and Applications

Find the area of a quadrilateral formed by the points A(1,2),B(4,6),C(7,5),D(2,1)A(1, 2), B(4, 6), C(7, 5), D(2, 1)A(1,2),B(4,6),C(7,5),D(2,1).

Answer:
Divide the quadrilateral into two triangles:
Triangle 1: A(1,2),B(4,6),D(2,1)A(1, 2), B(4, 6), D(2, 1)A(1,2),B(4,6),D(2,1)
Triangle 2: B(4,6),C(7,5),D(2,1)B(4, 6), C(7, 5), D(2, 1)B(4,6),C(7,5),D(2,1)
Using the area formula for each triangle and summing them up, you get the area.

Find the equation of the line passing through the point (3,5)(3, 5)(3,5) and parallel to the line 2x−3y=62x – 3y = 62x−3y=6.

Answer:
The slope of the given line is 23\frac{2}{3}32.
Using the point-slope form:
y−5=23(x−3)y – 5 = \frac{2}{3}(x – 3)y−5=32(x−3)
y−5=23x−2y – 5 = \frac{2}{3}x – 2y−5=32x−2
y=23x+3y = \frac{2}{3}x + 3y=32x+3.
So, the equation is y=23x+3y = \frac{2}{3}x + 3y=32x+3.

Section 11: Distance and Midpoint Theorem

Find the distance between the points P(3,4)P(3, 4)P(3,4) and Q(6,8)Q(6, 8)Q(6,8).

Answer:
Using the distance formula:
d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}d=(x2−x1)2+(y2−y1)2
=(6−3)2+(8−4)2=32+42=9+16=25=5= \sqrt{(6 – 3)^2 + (8 – 4)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5=(6−3)2+(8−4)2=32+42=9+16=25=5.
So, the distance is 5 units.

Find the midpoint of the segment joining A(−5,3)A(-5, 3)A(−5,3) and B(1,−2)B(1, -2)B(1,−2).

Answer:
Midpoint formula:
(x1+x22,y1+y22)\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)(2×1+x2,2y1+y2)
=(−5+12,3+(−2)2)=(−42,12)=(−2,12)= \left( \frac{-5 + 1}{2}, \frac{3 + (-2)}{2} \right) = \left( \frac{-4}{2}, \frac{1}{2} \right) = (-2, \frac{1}{2})=(2−5+1,23+(−2))=(2−4,21)=(−2,21).
So, the midpoint is (−2,12)(-2, \frac{1}{2})(−2,21).

What is the distance between the points A(0,0)A(0, 0)A(0,0) and B(5,12)B(5, 12)B(5,12)?

Answer:
Using the distance formula:
d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}d=(x2−x1)2+(y2−y1)2
=(5−0)2+(12−0)2=52+122=25+144=169=13= \sqrt{(5 – 0)^2 + (12 – 0)^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13=(5−0)2+(12−0)2=52+122=25+144=169=13.
So, the distance is 13 units.

If the midpoint of a segment is M(2,5)M(2, 5)M(2,5), and one end of the segment is A(1,3)A(1, 3)A(1,3), find the coordinates of the other endpoint B(x,y)B(x, y)B(x,y).

Answer:
Let B(x,y)B(x, y)B(x,y) be the other endpoint.
Midpoint formula:
(x1+x22,y1+y22)=M(2,5)\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = M(2, 5)(2×1+x2,2y1+y2)=M(2,5).
Substituting A(1,3)A(1, 3)A(1,3) and M(2,5)M(2, 5)M(2,5):
(1+x2,3+y2)=(2,5)\left( \frac{1 + x}{2}, \frac{3 + y}{2} \right) = (2, 5)(21+x,23+y)=(2,5).
Solving the equations:
1+x2=2⇒1+x=4⇒x=3\frac{1 + x}{2} = 2 \Rightarrow 1 + x = 4 \Rightarrow x = 321+x=2⇒1+x=4⇒x=3,
3+y2=5⇒3+y=10⇒y=7\frac{3 + y}{2} = 5 \Rightarrow 3 + y = 10 \Rightarrow y = 723+y=5⇒3+y=10⇒y=7.
So, B(3,7)B(3, 7)B(3,7).

Section 12: Equation of a Line in Different Forms

Find the equation of a line with slope 333 passing through the point (2,1)(2, 1)(2,1).

Answer:
Using the point-slope form:
y−y1=m(x−x1)y – y_1 = m(x – x_1)y−y1=m(x−x1)
y−1=3(x−2)y – 1 = 3(x – 2)y−1=3(x−2)
y−1=3x−6y – 1 = 3x – 6y−1=3x−6
y=3x−5y = 3x – 5y=3x−5.
So, the equation is y=3x−5y = 3x – 5y=3x−5.

Find the equation of the line passing through the points A(4,−1)A(4, -1)A(4,−1) and B(6,5)B(6, 5)B(6,5).

Answer:
First, calculate the slope:
m=5−(−1)6−4=62=3m = \frac{5 – (-1)}{6 – 4} = \frac{6}{2} = 3m=6−45−(−1)=26=3.
Using the point-slope form with point A(4,−1)A(4, -1)A(4,−1):
y−(−1)=3(x−4)y – (-1) = 3(x – 4)y−(−1)=3(x−4)
y+1=3x−12y + 1 = 3x – 12y+1=3x−12
y=3x−13y = 3x – 13y=3x−13.
So, the equation is y=3x−13y = 3x – 13y=3x−13.

Find the equation of the line through A(−2,3)A(-2, 3)A(−2,3) and perpendicular to the line 2x+3y=62x + 3y = 62x+3y=6.

Answer:
The slope of the given line is found by rewriting the equation in slope-intercept form:
2x+3y=6⇒y=−23x+22x + 3y = 6 \Rightarrow y = -\frac{2}{3}x + 22x+3y=6⇒y=−32x+2.
The slope of the perpendicular line is the negative reciprocal:
m=32m = \frac{3}{2}m=23.
Using the point-slope form:
y−3=32(x+2)y – 3 = \frac{3}{2}(x + 2)y−3=23(x+2)
y−3=32x+3y – 3 = \frac{3}{2}x + 3y−3=23x+3
y=32x+6y = \frac{3}{2}x + 6y=23x+6.
So, the equation is y=32x+6y = \frac{3}{2}x + 6y=23x+6.

Find the equation of the line through (1,4)(1, 4)(1,4) with slope −2-2−2.

Answer:
Using the point-slope form:
y−4=−2(x−1)y – 4 = -2(x – 1)y−4=−2(x−1)
y−4=−2x+2y – 4 = -2x + 2y−4=−2x+2
y=−2x+6y = -2x + 6y=−2x+6.
So, the equation is y=−2x+6y = -2x + 6y=−2x+6.

Section 13: Application of Coordinate Geometry

Find the area of a triangle formed by the points A(1,2),B(4,6),C(5,1)A(1, 2), B(4, 6), C(5, 1)A(1,2),B(4,6),C(5,1).

Answer:
Area of a triangle with vertices (x1,y1),(x2,y2),(x3,y3)(x_1, y_1), (x_2, y_2), (x_3, y_3)(x1,y1),(x2,y2),(x3,y3) is given by:Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|Area=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣

Substituting the coordinates:Area=12∣1(6−1)+4(1−2)+5(2−6)∣\text{Area} = \frac{1}{2} \left| 1(6 – 1) + 4(1 – 2) + 5(2 – 6) \right|Area=21∣1(6−1)+4(1−2)+5(2−6)∣

=12∣5−4−20∣=12×19=9.5= \frac{1}{2} \left| 5 – 4 – 20 \right| = \frac{1}{2} \times 19 = 9.5=21∣5−4−20∣=21×19=9.5.
So, the area of the triangle is 9.5 square units.

Find the distance between the point A(1,2)A(1, 2)A(1,2) and the line 3x−4y+5=03x – 4y + 5 = 03x−4y+5=0.

Answer:
The distance from a point (x1,y1)(x_1, y_1)(x1,y1) to a line Ax+By+C=0Ax + By + C = 0Ax+By+C=0 is given by:d=∣Ax1+By1+C∣A2+B2d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}d=A2+B2∣Ax1+By1+C∣

Substituting the values:
A=3,B=−4,C=5,x1=1,y1=2A = 3, B = -4, C = 5, x_1 = 1, y_1 = 2A=3,B=−4,C=5,x1=1,y1=2,d=∣3(1)−4(2)+5∣32+(−4)2=∣3−8+5∣9+16=∣0∣5=0d = \frac{|3(1) – 4(2) + 5|}{\sqrt{3^2 + (-4)^2}} = \frac{|3 – 8 + 5|}{\sqrt{9 + 16}} = \frac{|0|}{5} = 0d=32+(−4)2∣3(1)−4(2)+5∣=9+16∣3−8+5∣=5∣0∣=0

So, the point lies on the line and the distance is 0.

Find the equation of the line that passes through the point P(2,1)P(2, 1)P(2,1) and is parallel to the line y=−x+4y = -x + 4y=−x+4.

Answer:
The slope of the given line is −1-1−1, and since the line is parallel, it will have the same slope.
Using the point-slope form:
y−1=−1(x−2)y – 1 = -1(x – 2)y−1=−1(x−2)
y−1=−x+2y – 1 = -x + 2y−1=−x+2
y=−x+3y = -x + 3y=−x+3.
So, the equation is y=−x+3y = -x + 3y=−x+3.

Section 14: Miscellaneous Problems

Find the area of the quadrilateral formed by the points A(1,2),B(4,6),C(7,3),D(2,1)A(1, 2), B(4, 6), C(7, 3), D(2, 1)A(1,2),B(4,6),C(7,3),D(2,1).

Answer:
Divide the quadrilateral into two triangles:
Triangle 1: A(1,2),B(4,6),D(2,1)A(1, 2), B(4, 6), D(2, 1)A(1,2),B(4,6),D(2,1)
Triangle 2: B(4,6),C(7,3),D(2,1)B(4, 6), C(7, 3), D(2, 1)B(4,6),C(7,3),D(2,1)
Using the area formula for each triangle and summing the results, you will find the area.

Find the distance between the points (0,0)(0, 0)(0,0) and (x,y)(x, y)(x,y) if x=4x = 4x=4 and y=3y = 3y=3.

Answer:
Using the distance formula:
d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}d=(x2−x1)2+(y2−y1)2
d=(4−0)2+(3−0)2=42+32=16+9=25=5d = \sqrt{(4 – 0)^2 + (3 – 0)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5d=(4−0)2+(3−0)2=42+32=16+9=25=5.
So, the distance is 5 units.

This concludes a significant portion of questions for Chapter 3.

Also Read: Class 9 Chapter 2: Polynomials Worksheet for Exams

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