Question

What is the equation of the line in slope-intercept form that is perpendicular to the line y=3/4x-2 and passes

through the point (-12, 10)?
Oy=-4/3-6
O y=-4/3x + 6
O y = 4/3x + 26
O y = 4/3x +10

Answer 1

Answer: y=-(4/3)*x-6

Explanation:

The equation of any straight line is y=a*x+b (1).

So we have to find the coefficients a and b and substitute them to the equation (1).

If the required line is perpendicular to y= (3/4)*x-2 it means that

a= -(4/3) (we have to inverse the fraction 3/4 and put the opposite sign after that. 3/4 has the sign + in front of it so we have to put sign -)

So the equation of required line is y= -(4/3) *x+b .

Now we have to find b. To do that pls remember that the point (-12;10) belongs to the required line y= -(4/3) *x+b . That means:

10=-(4/3)*(-12)+b => 10=16+b => b=-6

So substitute b in equation (1) and get:

y=-(4/3)*x-6