Equation of a Line Exe-14C Class-10 Concise ICSE Maths Solution Ch-14. In this article you would learn concept of x and y intercept in equation of straight line. We Provide Step by Step Solutions / Answer of questions for Selina Concise Maths. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Equation of a Line Exe-14C Class-10 Concise ICSE Maths Solution Ch-14
| Board | ICSE |
| Publications | Selina |
| Subject | Maths |
| Class | 10th |
| Chapter-14 | Equation of a Line |
| Writer | R.K. Bansal |
| Exe-14C | x and y intercept in equation of straight line |
| Edition | 2025-2026 |
x and y intercept in equation of straight line
Equation of a Line Exe-14C Class-10 Concise ICSE Maths Solution Ch-14.
Que-1: Find the equation of line whose :
y-intercept = 2 and slope = 3,
Ans: Given, y-intercept = c = 2 and slope = m = 3.
Substituting the values of c and m in the equation y = mx + c, we get,
y = 3x + 2, which is the required equation.
Que-2: Find the equation of a line whose :
y-intercept = -1 and inclination = 45°
Ans: Given, y-intercept = c = −1 and inclination = 45°.
Slope = m = tan 45° = 1
Substituting the values of c and m in the equation y = mx + c, we get,
y = x – 1, which is the required equation.
Que-3: Find the equation of the line whose slope is -4/3 and which passes through (-3, 4).
Ans: Given, slope = -4/3
The equation passes through (−3, 4) = (x1, y1)
Substituting the values in y – y1 = m (x − x1), we get,
y-4=-4/3(x+3)
3y – 12 = −4x – 12
4x + 3y = 0, which is the required equation.
Que-4: Find the equation of a line which passes through (5, 4) and makes an angle of 60° with the positive direction of the x-axis.
Ans: The line passes through the point (5, 4) and angle of inclination = 60°
slope (m) = tan 60° = √3
Equation of line
y – y1 = m (x – x1)
and ⇒ y – 4 = √3 (x – 5)
hence ⇒ y – 4 = √3 x – 5√3
so ⇒ y = √3 x + 4 – 5√3
Que-5: Find the equation of the line passing through:
(i) (0, 1) and (1, 2)
(ii) (-1, -4) and (3, 0)
Ans:
(i)
Let (0, 1) = (x1, y1) and (1, 2) = (x2, y2)
∴ Slope of the line = 2-11-0=1
The required equation of the line is given by:
y – y1 = m(x – x1)
y – 1 = 1(x – 0)
y – 1 = x
y = x + 1
(ii)
Let (−1, −4) = (x1, y1) and (3, 0) = (x2, y2)
∴ Slope of the line = 0+43+1=44=1
The required equation of the line is given by:
y − y1 = m(x − x1)
y + 4 = 1(x + 1)
y + 4 = x + 1
y = x − 3
Que-6: The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find :
(i) The gradient of PQ and
(ii) The equation of PQ,
(iii) The co-ordinates of the point where PQ intersects the x-axis.
Ans: Given, co-ordinates of two points P and Q are (2, 6) and (–3, 5) respectively.
(i)
Gradient of PQ = 5-6/-3-2=-1/-5=1/5
(ii)
The equation of the line PQ is given by:
y − y1 = m(x − x1)
y-6=1/5(x-2)
5y − 30 = x − 2
5y = x + 28
(iii)
Let the line PQ intersects the x-axis at point A(x, 0).
Putting y = 0 in the equation of the line PQ, we get,
0 = x + 28
x = −28
Thus, the co-ordinates of the point where PQ intersects the x-axis are A(−28, 0).
Que-7: The co-ordinates of two points A and B are (-3, 4) and (2, -1). Find :
(i) the equation of AB
(ii) the co-ordinates of the point where the line AB intersects they- axis.
Ans:
(i)
Given, co-ordinates of two points A and B are (–3, 4) and (2, –1).
Slope = -1-4/2+3=-5/5=-1
The equation of the line AB is given by:
y − y1 = m(x − x1)
y + 1 = −1(x − 2)
y + 1 = −x + 2
x + y = 1
(ii)
Let the line AB intersects the y-axis at point (0, y).
Putting x = 0 in the equation of the line, we get,
0 + y = 1
y = 1
Thus, the co-ordinates of the point where the line AB intersects the y-axis are (0, 1).
Que-8: The figure given below shows two straight lines AB and CD intersecting each other at point P (3, 4). Find the equations of AB and CD.
Ans: Two lines AB and CD intersect each other at P (3, 4)
AB inclined at angle of 45° and CD at angle of 60° with the x-axis.
Slope of AB = tan 45° = 1
and slope of CD = tan 60° = √3
Now, equation of line AB will be
y – y1 = m (x – x1)
and ⇒ y – 4 = 1 (x – 3)
⇒ y – 4 = x – 3
so ⇒ y = x – 3 + 4
therefore ⇒ y = x + 1
(ii) Equation of CD will be :
y – y1 = m (x – x1)
and ⇒ y – 4 = √3 (x – 3)
so ⇒ y – 4 = √3 x – 3√3
therefore ⇒ y = √3 x – 3√3 + 4
⇒ y = √3 x + 4 – 3√3
Que-9: In ΔABC, A (3, 5), B (7, 8) and C (1, -10). Find the equation of the median through A.
Ans: AD is median
D is mid point of BC
Equation of AD
y – y1 = m (x – x1)
and ⇒ y + 1 = -6 (x – 4)
so ⇒ y + 1 = -6x + 24
hence ⇒ y + 6x = -1 + 24
⇒ 6x + y = 23
Que-10: The following figure shows a parallelogram ABCD whose side AB is parallel to the x-axis, ∠A = 60° and vertex, C = (7, 5). Find the equations of BC and CD.
Ans: ABCD is a ||gm in which AB = CD || x-axis
∠A = 60° and C (7, 5)
(i) CD || AB || x-axis ,
Equation of CD will be
y – y1 = m (x – x1)
and ⇒ y – 5 = 0 (x – 7)
so ⇒ y – 5 = 0
hence ⇒ y = 5
BC || AD
Slope of BC = tan 60° = √3
Equation of BC will be
y – y1 = m (x – x1)
and ⇒ y – 6 = √3 (x – 7)
so ⇒ y – 6 = √3 x – 7√3
therefore ⇒ y = √3 x + 6 – 7√3
Since, CD || AB and AB || x-axis, slope of CD = Slope of AB = 0
Equation of the line CD is given by:
y − y1 = m(x − x1)
y − 5 = 0(x − 7)
y = 5
Que-11: Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x – y = 4.
Ans: Point of intersection of two lines
x + 2y = 7 ….(i)
x – y = 4 ….(ii)
Subtracting, we get
3y = 3
y =1
Substituting, the value of y in (ii)
x – 1 = 4
⇒ x = 4 + 1 = 5
Point of intersection is (5, 1)
Slope of the line passing through origin (0, 0) and (5, 1)
Equation of line will be
y – y1 = m (x – x1)
y – 0 = 1/5 (x – 0)
5y = x
x – 5y = 0
Que-12: In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A. Also, find the equation of the line through vertex B and parallel to AC.
Ans: Given the co-ordinates of vertices A, B and C of a triangle ABC are (4, 7), (–2, 3) and (0, 1) respectively.
Let AD be the median through vertex A.
Co-ordinates of the point D are
(-2+0/2,3+1/2)
(–1, 2)
∴ Slope of AD = 2-7/-1-4=-5/-5=1
The equation of the median AD is given by:
y − y1 = m(x − x1)
y − 2 = 1(x + 1)
y − 2 = x + 1
y = x + 3
The slope of the line which is parallel to line AC will be equal to the slope of AC.
Slope of AC = 1-7/0-4=-6/-4=3/2
The equation of the line which is parallel to AC and passes through B is given by:
y-3=3/2(x+2)
2y − 6 = 3x + 6
2y = 3x + 12
Que-13: A, B and C have co-ordinates (0, 3), (4, 4) and (8, 0) respectively. Find the equation of the line through A and perpendicular to BC.
Ans: Slope of line through A, perpendicular to BC = -(-1) = 1
Now, the equation of line through A (0, 3) is
y – y1 = m (x – x1)
y – 3 = 1 (x – 0)
and ⇒ y – 3 = x
⇒ y = x + 3
Que-14: Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3).
Ans: Slope of the line joining the points (1, 4) and (2, 3)
Slope of line perpendicular to the above line = 1
Equation of line passing through (-1, 2)
y – y1 = m (x – x1)
⇒ y – 2 = 1 [x -(-1)]
and ⇒ y – 2 = x + 1
so y = x + 1 + 2
therefore ⇒ y = x + 3
Que-15: Find the equation of the line, whose :
(i) x-intercept = 5 and y-intercept = 3
(ii) x-intercept = -4 and y-intercept = 6
(iii) x-intercept = -8 and y-intercept = -4
(iv) x-intercept = 3 and y-intercept = -6
Ans:
(i)
When x-intercept = 5, then point will be (5, 0)
and when y-intercept = 3, then point will be (0, 3)
Let (x1, y1) = (5, 0) and (x2, y2) = (0, 3)
Slope = 3-0/0-5=-3/5
The required equation is:
y – y1 = m(x – x1)
y-0=-3/5(x-5)
5y = –3x + 15
3x + 5y = 15
(ii)
When x-intercept = −4, corresponding point on x-axis is (−4, 0)
When y-intercept = 6, corresponding point on y-axis is (0, 6).
Let (x1, y1) = (−4, 0) and (x2, y2) = (0, 6)
Slope = 6-0/0+4=6/4=3/2
The required equation is:
y – y1 = m(x – x1)
y-0=3/2(x+4)
2y = 3x + 12
(iii)
When x-intercept = −8, corresponding point on x-axis is (−8, 0)
When y-intercept = −4, corresponding point on y-axis is (0, −4).
Let (x1, y1) = (−8, 0) and (x2, y2) = (0, −4)
Slope = -4-0/0+8=-4/8=-1/2
The required equation is:
y − y1 = m(x − x1)
y-0=-1/2(x+8)
2y = −x − 8
x + 2y + 8 = 0
Que-16: Find the equation of the line whose slope is -5/6 and x-intercept is 6.
Ans: Since, x-intercept is 6, so the corresponding point on x-axis is (6, 0).
Slope = m = -5/6
Required equation of the line is given by:
y – y1 = m(x – x1)
y-0=-5/6(x-6)
6y = –5x + 30
5x + 6y = 30
h5>Que-17: Find the equation of the line with x-intercept 5 and a point on it (-3, 2).
Ans: x-intercept of the line = 5
Point = (5, 0)
Slope of the line passing through the point (-3, 2)
∴ Slope of the line = 2-0/-3-5=2/-8=-1/4
Required equation of the line is given by:
y – y1 = m(x – x1)
y-0=-1/4(x-5)
4y = –x + 5
x + 4y = 5
Que-18: Find the equation of the line through (1, 3) and making an intercept of 5 on the y- axis.
Ans: The line makes y-intercept = 5
Point = (0, 5)
Slope of the line passing through the point (1, 3) and (0, 5)
Slope of the line = 5-3/0-1=2/-1=-2
Required equation of the line is given by:
y – y1 = m (x – x1)
and ⇒ y – 3 = -2 (x – 1)
so ⇒ y – 3 = -2x + 2
hence ⇒ 2x + y = 2 + 3
⇒ 2x + y = 5
Que-19: Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axes.
Ans: Slope of line AB = tan 45° = 1
Equation passing through the point (-2, 0)
y – y1 = m (x – x1)
⇒ y – 0 = 1 (x + 2)
and⇒ y = x + 2
⇒ x – y + 2 = 0
Slope of line CD = tan (-45°) = -1
Equation passing through the point (-2, 0)
y – y1 = m (x – x1)
⇒ y – 0 = -1 (x + 2)
and ⇒ y = -x – 2
so ⇒ y + x + 2 = 0
hence⇒ x + y + 2 = 0
Que-20: The line through P (5, 3) intersects y axis at Q.
(i) Write the slope of the line.
(ii) Write the equation of the line.
(iii) Find the co-ordinates of Q.
Ans:
(i)
Here θ = 45°
So, slope of the line = tanθ = tan 45° = 1
(ii)
Equation of the line through P and Q is
y – 3 = 1 (x – 5)
y – x + 2 = 0
x – y = 2
(iii)
Given that the line intersects with the y-axis, x = 0
Thus, substituting x = 0 in the equation x – y = 2
We have, 0 – y = 2
y = −2
Thus, the coordinates point of intersection Q are q(0, −2).
Que-21: Write down the equation of the line whose gradient is -2/5 and which passes through point P, where P divides the line segment joining A (4, -8) and B (12, 0) in the ratio 3 : 1.
Ans: Given, P divides the line segment joining A(4, −8) and B(12, 0) in the ratio 3 : 1.
Co-ordinates of point P are
(3×12+1×4/3+1,3×0+1×(-8)/3+1)
= (36+4/4,0-8/4)
= (40/4,-8/4)
= (10, −2)
Slope = m = -2/5 …(Given)
Thus, the required equation of the line is
y − y1 = m(x − x1)
y+2=-2/5(x-10)
5y + 10 = −2x + 20
2x + 5y = 10
Que-22: A (1, 4), B (3, 2) and C (7, 5) are vertices of a triangle ABC. Find :
(i) the co-ordinates of the centroid of ΔABC.
(ii) the equation of a line, through the centroid and parallel to AB. [2002]
Ans:
(i)
Co-ordinates of the centroid –
x=x1+x2+x3/3
= 1+3+7/3
= 11/3
y=y1+y2+y3/3
= 4+2+5/3
= 11/3
The co-ordinates of the centroid of triangle ABC are (11/3 , 11/3)
(ii)
Slope of the line parallel to AB = Slope of AB = −1
Thus, the required equation of the line is
y – y1 = m(x – x1)
y-11/3=-1(x-11/3)
y-11/3=-x+11/3
y+x=11/3+11/3
y+x=11+11/3
= 22/3
3y + 3x = 22
Que-23: A (7, -1), B (4, 1) and C (-3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point F in AC; such that AP : CP = 2 : 3.
Ans: P divides AC in the ratio of 2 : 3
∴ Co-ordinates of P will be
(m1x2+m2x1/m1+m2,m1y2+m2y1/m1+m2)
(2(-3)+3(7)/2+3,2×4+3×(-1)/2+3)
= (-6+21/5,8-3/5)
= (15/5,5/5)
= (3, 1)
∴ Slope of line passing through B and P
= y2-y1/x2-x1
= 1-1/3-4
= 0/-1
= 0
∴ Equation of the required line is given by y – y1 = m(x – x1)
y – 1 = 0(x – 4)
y – 1 = 0
y = 1
–: Equation of a Line Exe-14C Class-10 Concise ICSE Maths Solution Ch-14. :–
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