Question

A line is such that its segment between the lines 4x + 3y - 21 = 0 and
10x + y - 59 = 0 is bisected at the point (4, 6). Find its equation.​

Answer 1

Answer:

The equation of new line  is  y = - 2.08 x + 14.28

Step-by-step explanation:

Given as :

The given lines are

4 x + 3 y = 21          ....1

10 x + y = 59           ......2

Solving the equations

3 (10 x + y) - (4 x + 3 y) = 3 × 59 - 21

Or, (30 x - 4 x) + (3 y - 3 y) = 177 - 21

Or, 16 x + 0 = 156

Or, x = $\dfrac{156}{16}$

i.e  x = 9.75

Put the value of x in eq 1

4 × 9.75 + 3 y = 21  

Or, 3 y = 21 - 39

Or, 3 y = - 18

Or, y = $\dfrac{-18}{3}$

i.e  y = - 6

So, The intersection points are ( 9.75 , - 6 )

Again

The other line bisect at point ( 4 , 6 )

So, The equation of new line thus formed

y - $y_1$  = m ( x - $x_1$ )

where m is the slope

m = $\dfrac{y_2-y_1}{x_2-x_1}$

Or, m = $\dfrac{6+6}{4-9.75}$

So, slope = m = - 2.08

The equation of new line

y + 6  = ( - 2.08) ( x - 9.75 )

Or, y + 6 = - 2.08 x + 20.28

Or, y = - 2.08 x + 20.28 - 6

∴  y = - 2.08 x  +  14.28

Hence,  The equation of new line  is  y = - 2.08 x + 14.28  Answer