Question

if (-4,5) is a vertex of a square and one of its diagonal is 7x-y+8=0 .find the equation of other diagonal

Answer 1

Thank you for asking this question.

The options for this question are missing. Following are the missing options:

A) x+3y=21

B) 2x−3y=7

C) x+7y=31

D) 2x+3y=21

The correct answer is   B) 2x−3y=7

Reason:

Equation of perpendicular line to 7x−y+8=0 is x+7y=h which is passes through (−4, 5)  

h = 31  

So, equation of another diagonal is x+7y = 31

Answer 2

x + 7y = 31

Step-by-step explanation:

Given:

The vertex of a square is (-4, 5)

And the equation of one diagonal is 7x - y + 8 = 0

Let the equation of the other diagonal be y = mx + c where m is the slope of the line and c is the y- intercept.

Since this line passes through (- 4, 5)

$5 = -4m +c$-------------(1)

We also write the given equation in standard form $y = mx + c$

where m is the slope of the line and c is the y intercept.

$7x - y + 8 = 0$

$y = 7x + 8$

So, the slope of the above line is m = 7

We know that the perpendicular line slope is = $-\frac{1}{m}$

So, the slope of the perpendicular line is $-\frac{1}{7}$

Putting the value of perpendicular slope in equation 1 for value of c.

$5 = -4\times (-\frac{1}{7}) + c$

$5 = \frac{4}{7} + c$

$c = 5 - \frac{4}{7}$

$c =\frac{7\times 5 - 4}{7} \\c= \frac{35-4}{7} \\c=\frac{31}{7}$

Now, we substitute value of perpendicular slope and value of c in below equation.

$y=mx+c$

$y = (-\frac{1}{7})x + \frac{31}{7}$

$y=-\frac{x}{7} + \frac{31}{7}$

$y = \frac{-x+31}{7}$

$7y=-x+31$

$x+7y=31$

Therefore, the equation of other diagonal is $x+7y=31$