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Lars:

Points P and Q are two of the vertices of a right triangle. They are the endpoints of the hypotenuse PQ¯¯¯¯¯ of the triangle. The third vertex lies in the second quadrant. Plot a point at the location of the third vertex.

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Lars:

Is my answer -2, -2? If so where on this coordinate plane am I supposed to plot it? I'm confused

Lars:

Did I correctly plot it?

K1NGofPadlet:
No, the answer is not -2, -2.
To find the solution, you require to first find the x-intercept and y-intercept of the two lines.
For line p, the x-intercept is -2 and the y-intercept is 3. So the equation of line p is y = (-3/2)x + 3.
For line q, the x-intercept is 3 and the y-intercept is -2. So the equation of line q is y = (2/3)x - 2.
To find the point of intersection, you require to solve for x and y in the system of equations:
y = (-3/2)x + 3
y = (2/3)x - 2
Setting the two equations identically tantamount to each other and solving for x, we get:
(-3/2)x + 3 = (2/3)x - 2
(-3/2)x - (2/3)x = -2 - 3
(-9/6)x - (4/6)x = -5
(-13/6)x = -5
x = 30/13
Then, plugging in x = 30/13 into either of the equations and solving for y, we get:
y = (-3/2)(30/13) + 3 = -9/13
So the point of intersection is (30/13, -9/13). Plot this point on the coordinate plane to show the solution.

@lars wrote:

Is my answer -2, -2? If so where on this coordinate plane am I supposed to plot it? I'm confused

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