Name another point IN QUAD II on the line that is perpendicular to the given line and passing through the point (-2, 1).

Question

note: in the question, there is a picture of a graph with a line going through 2 points

point 1 is (3,0)

point 2 is (0,-2)

Answer 1

Remember the whole graph has 4 quadrants

Quad II=(-x,y) Quad I= (x,y)

Quad III=(-x,-y). Quad IV=(x,-y)

The question states that the line has to be perpendicular.

First you have to find the slope of the two points that the line passes through. Use those coordinates to find the slope.

The formula for slope is

y2-y1 (0--2)

____=______=2/3=m

x2-x1. (3-0)

Then you need to use the point slope formula

(y-y1)=m(x-x1)

Since the value of y is positive it's subtraction.

since the value of x is negative it's positive.

y-1=2/3(x+2)

+1 +1

____________

y=2/3 (x+3)

Then you add one to each side of the equation

y=2/3x+2

That's the solution after distribution.

ANS: y=2/3x+2

Answer 2

Calculate the formula for L1 the line through (0,-1) and (3,0).

Calculate the formula for the line L2 through (-2,1) and (0,-2) which is perpendicular to L1 at (0,-2)

Then choose another point on L2 say at x=-4, and it will pass through (-4,4).

Answer 3

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