Question

Equation of the straight line whose slope is 3 and passing through the point of intersection of the lines 3x + 4y = 7 and
x − y + 2 = 0 is ____.

• A. $21x-7y+16=0$
• B. $9x-3y+14=0$
• C. $21x-7y+12=0$
• D. $9x-3y+5=0$

Answer 1

The correct option is A $21x-7y+16=0$

On solving the two equations
3x + 4y = 7
x − y + 2 = 0,

3x+4y = 7
3x-3y = -6
(on multiplying second equation by 3)
Solving the above 2 we get,
7y = 13
$y=\frac{13}{7}$
Substituting the value of y in x-y+2 = 0 and solving for x, we get $x=\frac{-1}{7}$

So, we get the point of intersection as $(\frac{-1}{7},\frac{13}{7}$)

Slope of the required equation $m=3$

Therefore, using point-slope form

$(y-y_1)=m(x-x_1)$

$y-\frac{13}{7}=3(x-\frac{-1}{7})$

$7y-13=21x+3$

$21x-7y+16=0$