Question

On a coordinate plane, line M N goes through points (2, 3) and (negative 3, 2). Point K is at (3, negative 3). Which point could be on the line that is perpendicular to Line M N and passes through point K? (0, −12) (2, 2) (4, 8) (5, 13)

Answer 1

MN goes through

• (2,3)
• (-3,2)

Slope:-

• m=2-3/-3-2=-1/-5=1/5

Slope of perpendicular line:-

• -5

Passes through k(3,-3)

Equation in point slope form

• y-y1=m(x-x1)
• y+3=-5(x-3)
• y+3=-5x+15
• y=-5x+12

Check one by one

• 0=-5(-12)+12=60+12=72
• 2=-5(2)+12=-10+12=2

Yes option B is correct

Answer 2

Answer:

(2, 2)

Step-by-step explanation:

First find the slope of the line MN, using the two given points:

$\sf let\:(x_1,y_1)=(2,3)$

$\sf let\:(x_2,y_2)=(-3,2)$

$\implies \sf slope\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{2-3}{-3-2}=\dfrac15$

If two lines are perpendicular to each other, the product of their slopes will be -1.

Therefore, the slope (m) of the line perpendicular to MN is:

$\implies \dfrac15 \times m=-1$

$\implies m=-5$

Using the point-slope form of the linear equation with point K (3, -3), we can construct the linear equation of the line perpendicular to MN:

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

$\implies \sf y-(-3)=-5(x-3)$

$\implies \sf y=-5x+12$

Now we have the equation of the line perpendicular to MN, we can input the x-values of the solution options to see which once is correct:

$x=0\implies \sf y=-5(0)+12=12\implies (0,12)$

$x=2\implies \sf y=-5(2)+12=2\implies (2,2)$

$x=4\implies \sf y=-5(4)+12=12\implies (4,-8)$

$x=5\implies \sf y=-5(5)+12=-13\implies (5,-13)$

Therefore, the point that is on the line that is perpendicular to MN and passes through point K is (2, 2)