Question

Find the point of intersection of the line joining points (– 3, 7) (2, – 4) and (4, 6) (–5, – 7). Also, find the point of intersection of these lines and also their intersection with the axis.

Answer 1

Line 1: Joining points x₁ y₁ = (−3, 7) and x₂ y₂ = (2, −4)

Equation of line joining 2 points by 2 point formula is given by

`(y - y_1)/(x - x_1) = (y_2 - y_1)/(x_2 - x_1)`

∴ `(y - 7)/(x - (-3)) = (-4 - 7)/(2 - (-3))`

`(y - 7)/(x + 3) = (-11)/(2 + 3)`

∴ `(y - 7)/(x + 3) = (-11)/5`

Cross multiplying, we get

5(y − 7) = −11(x + 3)

5y − 35 = −11x − 33

Transposing the variables, we get

11x + 5y = 35 – 33 = 2

11x + 5y = 2 ......(line 1)

Similarly, we should find out equation of second line

Line 2: Joining points x₁ y₂ = (4, 6) and x₂ y₂ = (−5, −7)

`(y - y_1)/(x - x_1) = (y_2 - y_1)/(x_2 - x_1)`

∴ `(y - 6)/(x - 4) = (-7 - 6)/(-5 - 4)`

∴ `(y - 6)/(x - 4) = (-13)/(-9) = 13/9`

∴ 9y – 54 = 13x – 52

∴ 9y – 13x = 2    ...(line 2)

For finding point of intersection, we need to solve the 2 line equation to find a point that will satisfy both the line equations.

∴ Solving for x and y from line 1 and line 2 as below

11x + 5y = 2

⇒ multiply both sides by 13,

11 × 13x + 5 × 13y = 26      ...(3)

9y – 13x = 2 

⇒ multiply both sides by 11

9 × 11y – 13 × 11x = 22      ...(4)

i.e.                            65y + 143x = 26  ...(3)
                             + 99y − 143x = 22  ...(4) 
Adding (3) and (4), 65y + 99y + 0 = 26 + 22 

∴ 164y = 48

∴ y = `48/164 = 12/41`

Substituting this value of y in line 1, we get

11x + 5y = 2

`11x + 5 xx 12/41` = 2

11x = `2 - 60/41 = (82- 60)/41 = 22/41`

∴ x = `2/41`

∴ Point of intersection is `(2/41, 12/41)`

To find point of intersection of the lines with the axis, we should substitute values and check

For line 1,

11x + 5y = 2

Point of intersection of line with x-axis, i.e., y coordinate is ‘0’

∴ put y = 0 in above equation

∴ 11x – 5 × 0 = 2

∴ 11x + 0 = 2

∴ x = `2/11`

∴ Points is `(2/11, 0)`

Similarly, Point of intersection of line with y-axis is when x-coordinate becomes ‘0’

∴ put x = 0 in above equation

∴ 11 × 0 + 5y = 2

∴ 0 + 5y = 2

y = `2/5`

∴ Points is `(0, 2/5)`

Similarly for line 2,

9y – 13x = 2

For finding x intercept, i.e., the point where line meets x-axis, we know that y coordinate becomes ‘0’

∴ Substituting y = 0 in above equation, we get

9 × 0 – 13x = 2

∴ 0 – 13x = 2

∴ x = `(-2)/13`

∴ Point: `((-2)/13, 0)`

Similarly for y-intercept, x-coordinate becomes ‘0’.

∴ Substituting for x = 0 in above equation, we get

9y – 13 × 0 = 2

9y – 0 = 2

9y = 2

y = `2/9`

Point `(0, 2/9)`