ML Aggarwal Solution Class 10 Chapter 12 Equation of a Straight Line Exercise 12.1
Exercise 12.1
Question 1
Find the slope of a line whose inclination is
(i) 45°
(ii) 30°
Sol :
(i) tan 45° = 1
(ii) tan 30°=1√3
Question 2
Find the inclination of a line whose gradient is
(i) 1
(ii) √3
Sol :
(i) tan θ = 1 ⇒ θ = 45°
(ii) tan θ = √3 ⇒ θ = 60°
Question 3
Find the equation of a straight line parallel 1 to x-axis which is at a distance
(i) 2 units above it
(ii) 3 units below it.
Sol :
(i) A line which is parallel to x-axis is y = a
⇒ y = 2
⇒ y – 2 = 0
(ii) A line which is parallel to x-axis is y = a
⇒ y = -3
⇒ y + 3 = 0
Question 4
Find the equation of a straight line parallel to y-axis which is at a distance of:
(i) 3 units to the right
(ii) 2 units to the left.
Sol :
(i) The equation of line parallel to y-axis is at a distance of 3 units to the right is x = 3 ⇒ x – 3 = 0
(ii) The equation of line parallel to y-axis at a distance of 2 units to the left is x = -2 ⇒ x + 2 = 0
Question 5
Find the equation of a straight line parallel to y-axis and passing through the point ( – 3, 5).
Sol :
The equation of the line parallel to y-axis passing through ( – 3, 5) to x = -3
⇒ x + 3 = 0
Question 6
Find the equation of the a line whose
(i) slope = 3, y-intercept = – 5
(iii) gradient =√3,y -intercept =−43
(iv) inclination = 30°,y-intercept = 2
Sol :
Equation of a line whose slope and y-intercept is given is
y = mx + c
where m is the slope and c is the y-intercept
(i) y = mx + c
⇒ y = 3x + (-5)
⇒ y = 3x – 5
(ii) y=mx+c⇒y=−27x+3
(iii) y=mx+c
⇒y=√3x+(−43)
⇒y=√3x−43
⇒3y=3√3x−4
⇒3√3x−3y−4=0
(iv) Inclination =30∘
∴ slope =tan30∘=1√3
∴ Equation y=mx+c⇒y=1√3x+2
⇒√3y=x+2√3
⇒x−√3y+2√3=0
Question 7
Find the slope and y-intercept of the following lines:
(i) x – 2y – 1 = 0
(ii) 4x – 5y – 9 = – 0
(iii) 3x +5y + 7 = 0
(v) y – 3 = 0
(vi) x – 3 = 0
Sol :
We know that in the equation
y=mx+c, m is the slope and c is the y-intercept.
Now using this, we find,
⇒2y=x−1
⇒y=12x−12
Here slope =12 and y-intercept =−12
(ii) 4x−5y−9=0⇒4x−9=5y
⇒5y=4x−9
⇒y=45x−95
Here slope =45 and intercept =−95
(iii) 3x+5y+7=0
⇒5y=−3x−7
⇒y=−35x−75
Here slope =−35 and y-intercept =−75
(iv) x3+y4=1
⇒4x+3y=12
⇒3y=−4x+12
⇒y=−43x+123
⇒y=−43x+4
Here, =−43 and y-intercept =4
(v) y-3=0
⇒y=3⇒y=0,x+3
Here slope =0 and y-intercept =3
(vi) x-3=0
Here in this equation, slope cannot be defined and does not meet y-axis.
Question 8
The equation of the line PQ is 3y – 3x + 7 = 0
(i) Write down the slope of the line PQ.
(ii) Calculate the angle that the line PQ makes with the positive direction of x-axis.
Sol :
Equation of line PQ is 3y–3x+7=0
Writing in form of y=mx+c
⇒y=x−73
(i) Here slope=1
(ii) ∴ Angle which makes PQ with x-axis is Q
But tan.θ=1∴θ=45∘
Question 9
The given figure represents the line y = x + 1 and y = √3x – 1. Write down the angles which the lines make with the positive direction of the x-axis. Hence determine θ.
⇒tanθ=1
⇒θ=45∘
and slope of line y=√3x−1
m=√3⇒tanθ=√3
⇒θ=60∘
Now in Δ formed by the given two lines and x-axis.
Ext. angle=Sum of interior opposite angle.
⇒60∘=θ+45∘
⇒θ=60∘−45∘=15∘
Question 10
Find the value of p, given that the line y2=x−p passes through the point ( – 4, 4) (1992).
It passes through the points (-4, 4)
It will satisfy the equation
Hence, p=-6
Question 11
Given that (a, 2a) lies on the line y2=3x−6 find the value of a
Sol :
∵ Point (a, 2a) lies on the line
∴2a2=3(a)−6
⇒a=3a+6
-3a+a=-6
⇒−2a=−6
⇒a=−6−2
∴a=3
Question 12
The graph of the equation y = mx + c passes through the points (1, 4) and ( – 2, – 5). Determine the values of m and c.
Sol :
Equation of the line is y = mx + c
∴ it passes through the points (1, 4)
∴ 4 = m x 1 + c
⇒ 4 = m + c
⇒ m + c = 4 … (i)
Again it passes through the point (-2, -5)
∴ 5 = m (-2) + c
⇒ 5 = -2 m + c
⇒ 2m – c = 5 …(ii)
Adding (i) and (ii)
3m = 9
⇒ m = 3
Substituting the value of m in (i)
3 + c = 4
⇒ c = 4 – 3 = 1
Hence m = 3, c = 1
Question 13
Find the equation of the line passing through the point (2, – 5) and making an intercept of – 3 on the y-axis.
Sol :
∴ The line intersects y-axis making an intercept of -3
∴ the co-ordinates of point of intersection will be (0, -3)
=−3+50−2=2−2=−1
∴ Equation of the line will be,
y−y1=m(x−x1)
⇒y−(−5)=−1(x−2)
⇒y+5=−x+2
⇒x+y+5−2=0
⇒x+y+3=0
Question 14
Find the equation of a straight line passing through ( – 1, 2) and whose slope is 25
Sol :
Equation of the line will be
y−y1=m(x−x1)
y−2=25(x+1)
⇒ 5y – 10 = 2x + 2
⇒ 2x – 5y + 2 + 10 = 0
⇒ 2x – 5y + 12 = 0
Question 15
Find the equation of a straight line whose inclination is 60° and which passes through the point (0, – 3).
Sol :
The equation of line whose slope is wand passes through a given point is
Here m = tan 60° = √3 and point is (0, -3)
∴ y + 3 = √3 (x – 0)
⇒ y + 3 = √3x
⇒ √3x – y – 3 = 0
Question 16
Find the gradient of a line passing through the following pairs of points.
(i) (0, – 2), (3, 4)
(ii) (3, – 7), ( – 1, 8)
Given
(i) (0, -2), (3, 4)
(ii) (3, -7), (-1, 8)
(ii) m=8+7−1−3=15−4 ∴gradient=−154
Question 17
The coordinates of two points E and F are (0, 4) and (3, 7) respectively. Find :
(i) The gradient of EF
(ii) The equation of EF
(iii) The coordinates of the point where the line EF intersects the x-axis.
Sol :
Co-ordinates of points E (0, 4) and F (3, 7) are given, then
(i) The gradient of EF
(ii) Equation of line EF,
y−y1=m(x−x1)
⇒y−7=1(x−3)
⇒y−7=x−3
⇒x−y−3+7=0
⇒x−y+4=0
(iii) Co-ordinates of point of intersection of EF and the x-axis will be y=0
Substitutes the value y in the above equation
x-y+4=0
⇒x−0+4=0(∵y=0)
⇒x=−4
Hence co-ordinates are (-4,0)
Question 18
Find the intercepts made by the line 2x – 3y + 12 = 0 on the co-ordinate axis.
Sol :
Putting y = 0, we will get the intercept made on x-axis,
2x – 3y + 12 = 0
⇒ 2x – 3 × 0 + 12 = 0
⇒ 2x – 0 + 2 = 0
⇒ 2x = -12
⇒ x = -6
and putting x = 0, we get the intercepts made on y-axis,
2x – 3y + 12 = 0
⇒ 2 × 0 – 3y + 12 = 0
⇒ -3y = -12
Question 19
Find the equation of the line passing through the points P (5, 1) and Q (1, – 1). Hence, show that the points P, Q and R (11, 4) are collinear.
Sol :
The two given points are P (5, 1), Q(1, -1).
∴ Slope of the line (m)
Equation of the line,
y−y1=m(x−x1)
⇒y+1=12(x−1)
⇒2y+2=x−1
⇒x−2y−1−2=0...(i)
If point R(11,4) be on it, then it will satisfy it.
Now substituting the value of x and y in
(i) 11−2×4−3=11−8−3=11−11=0
∴ R satisfies it
Hence P, Q and R are collinear.
Question 20
Find the value of ‘a’ for which the following points A (a, 3), B (2,1) and C (5, a) are collinear. Hence find the equation of the line.
Sol :
Given That
A(a, 3), B (2, 1) and C (5, a) are collinear.
Slope of AB = Slope of BC
-6=(a-1)(2-a) (Cross-multiplication)
−6=2a−a2−2+a
−6=3a−a2−2
a2−3a+2−6=0
a2−3a−4=0
a2−4a+a−4=0
a(a-4)+1(a-4)=0
(a+1)(a-4)=0
a=-1, or a=4
a=-1 (∵does not satisfy the equation)
∴a=4
Slope of BC=a−15−2=4−13=33=1 m
Equation of BC ;(y-1)=1(x-2)
y−1=x−2⇒x−y=−1+2
x-y=1
Question 21
Use a graph paper for this question. The graph of a linear equation in x and y, passes through A ( – 1, – 1) and B (2, 5). From your graph, find the values of h and k, if the line passes through (h, 4) and (12, k). (2005)
Sol :
Points (h, 4) and (12, k) lie on the line passing
through A(-1, -1) and B(2, 5)
and k=2
Question 22
ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, – 4). Find
(i) the coordinates of A
(ii) the equation of the diagonal BD.
Sol :
Given that
ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4)
Coordinates of O=(5+22,8−42)=(3.5,2)
For the line AC
3.5=x+42
⇒x+4=7
⇒x=7−4=3
x=3, y=-3
2=y+72
⇒y+7=4
⇒y=4−7=−3
Thus, the coordinates of A are (3,-3)
(ii) Equation of diagonal BD is given by
y−8=−4−82−5(x−5)
⇒y−8=−12−3(x−5)
⇒y−8=4x−20
⇒4x−y−12=0
Question 23
In ∆ABC, A (3, 5), B (7, 8) and C (1, – 10). Find the equation of the median through A.
Sol :
AD is median
⇒ D is mid point of BC
i.e (4, -1)
slope of AD
m=y2−y1x2−x1=5+13−4=6−1=−6
∴ Equation of AD
y−y1=m(x−x1)
⇒y+1=−6(x−4)
⇒y+1=−6x+24
⇒y+6x=−1+24
⇒6x+y=23
Question 24
Find the equation of a line passing through the point ( – 2, 3) and having x-intercept 4 units. (2002)
Sol :
x-intercept = 4
∴ Co-ordinates of the point will be (4, 0)
Now slope of the line passing through the points (-2, 3) and (4, 0)
∴ Equation of the line will be
y−y1=m(x−x1)
⇒y−0=−12(x−4)
⇒2y=−x+4
⇒x+2y=4 or x+2y-4=0
Question 25
Find the equation of the line whose x-intercept is 6 and y-intercept is – 4.
Sol :
x-intercept = 6
∴ The line will pass through the point (6, 0)
y -intercept = -4 ⇒ c = -4
Now m=y2−y1x2−x1=−4−00−6=−4−6=23
∴ Equation of line will be
y=m x+c
⇒y=23x+(−4)=23x−4
⇒3y=2x−12⇒2x−3y=12
Question 26
Write down the equation of the line whose gradient is 12 and which passes through P where P divides the line segment joining A ( – 2, 6) and B (3, – 4) in the ratio 2 : 3. (2001)
=6−65=05=0
y=m1y2+m2y1m1+m2=2×(−4)+3(6)2+3
=−8+185=105=2
∴ Co-ordinates are (0,2)
Now slope (m) of the line passing through (0,2)=32
∴ Equation of the line will be
y−y1=m(x−x1)
⇒y−2=32(x−0)
⇒2y−4=3x
⇒3x−2y+4=0
Question 27
Find the equation of the line passing through the point (1, 4) and intersecting the line x – 2y – 11 = 0 on the y-axis.
Sol :
line x – 2y – 11 = 0 passes through y-axis
x = 0,
Now substituting the value of x in the equation x – 2y – 11 = 0
∴ Co-ordinates of point will be (0,−112)
Now slope of the line joining the points
(1,4) and (0,−112)
m=y2−y1x2−x1=−112−40−1=−192−1=192
and equation of the line will be
y−y1=m(x−x1)
⇒y+112=192(x−0)
⇒2y+11=19x
⇒19x−2y−11=0
Question 28
Find the equation of the straight line containing the point (3, 2) and making positive equal intercepts on axes.
=0−yx−0=−xx=−1(∵x=y)
∴ Equation of the line will be y−y1=m(x−x1)
⇒y−2=−1(x−3)
⇒y−2=−x+3
⇒x+y−2−3=0
⇒x+y−5=0
⇒x+y=5
Question 29
Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2) find:
(i) the coordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD. (2015)
Sol :
Three vertices of a parallelogram ABCD taken in order are
A (3, 6), B (5, 10) and C (3, 2)
(i) We need to find the co-ordinates of D
We know that the diagonals of a parallelogram bisect each other
Let (x, y) be the co-ordinates of D
=(3,4)
And, mid-point of diagonal BD=(5+x2,10+y2)
Thus, we have
5+x2=3 and 10+y2=4
⇒5+x=6 and 10+y=8
⇒x=1 and y=-2
∴ Coordinate of D=(1,-2)
(ii) Length of diagonal BD
=√(1−5)2+(−2−10)2=√(4)2+(−12)2
=√16+144=√160 units
(iii) Equation of the side joining A(3,6) and D(1,-2) is given by
x−33−1=y−66+2
⇒x−32=y−68
⇒4(x−3)=y−6
⇒4x−12=y−6
⇒4x−y=6
Thus, the equation of the side joining A(3,6) and D(1,-2) is 4 x-y=6
Question 30
A and B arc two points on the x-axis and y-axis respectively. P (2, – 3) is the mid point of AB. Find the
(i) the co-ordinates of A and B.
(ii) the slope of the line AB.
(iii) the equation of the line AB. (2010)
Sol :
Points A and B are on x-axis and y-axis respectively
Let co-ordinates of A be (X, O) and of B be (O, Y)
P (2, -3) is the midpoint of AB
⇒x=4,y=−6
(i) Hence co-ordinates of A are (4,0) and of B are (0,-6)
(ii) Slope of AB=y2−y1x2−y1
=−6−00−4=−6−4=32
(iii) Equation of AB will be y−y1=m(x−x1)
⇒y=(−3)=32(x−2) (∵P lies on it )
⇒y+3=32(x−2)
⇒2y+6=3x−6
⇒3x−2y=6+6
⇒3x−2y=12
Question 31
Find the equations of the diagonals of a rectangle whose sides are x = – 1, x = 2 , y = – 2 and y = 6.
These lines form a rectangle when they intersect at A, B, C, D respectively
Co-ordinates of A, B, C and D will be
(-1, -2), (2, -2), (2, 6) and (-1, 6) respectively.
AC and BD are its diagonals
(i) Slope of the diagonal AC
∴ Equation of AC will be
y−y1=m(x−x1)
=y+2=83(x+1)
⇒3y+6=8x+8
8x-3y+8-6=0
⇒8x−3y+2=0
(ii) Slope of BD=y2−y1x2−x2=6+2−1−2=8−3
∴ Equation BD will be y−y1=m(x−x1)
⇒y+2=−83(x−2)
3y+6=-8x+16
⇒8x+3y+6−16=0
⇒8x+3y−10=0
Question 32
Find the equation of a straight line passing through the origin and through the point of intersection of the lines 5x + 1y – 3 and 2x – 3y = 7
Sol :
5x + 7y = 3 …(i)
2x – 3 y = 7 …(ii)
Multiply (i) by 3 and (ii) by 7,
15x + 21y = 9
14x – 21y = 49
Adding we get,
Substituting the value of x in (i)
5×2+7y=3⇒10+7y=3
⇒7y=3−10⇒7y=−7⇒y=−1
∴ Point of intersection of lines is (2,-1)
Now slope of the line joining the points
(2,-1) and the origin (0,0)
m=y2−y1x2−x1=0+10−2=−12
Equation of line will be
y−y1=m(x−x1)⇒y−0=−12(x−0)
⇒2y=−x
⇒x+2y=0
Question 33
Point A (3, – 2) on reflection in the x-axis is mapped as A’ and point B on reflection in the y-axis is mapped onto B’ ( – 4, 3).
(i) Write down the co-ordinates of A’ and B.
(ii) Find the slope of the line A’B, hence find its inclination.
Sol :
A’ is the image of A (3, -2) on reflection in the x-axis.
∴ Co-ordinates of A’ will be (3, 2)
Again B’ (- 4, 3) in the image of A’, when reflected in the y-axis
∴ Co-ordinates of B will be (4, 3)
(ii) Slope of the line joining, the points A’ (3, 2) and B (4, 3)
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