HELP ME! WILL MARB BRAINLIEST FOR THE BEST ANSWER. PLEASE SHOW FULL SOLUTIONS. THANK YOU AND GOD BLESS. 2. Given triangle PQR where P(4,1),Q(a, 2a) and R(3,4). Solve for a if the equation of the altitude to side QR is y = −x + 6. You may want to draw a diagram.

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  a = 1

Step-by-step explanation:

The altitude is perpendicular to line QR, so point Q lies on a line through R that is perpendicular to the given altitude line. The given altitude line has a slope of -1. This means line QR will have a slope of 1, so the line QR in point-slope form is ...

  y -k = m(x -h) . . . . . line with slope m through point (h, k)

  y -4 = 1(x -3)

Since point Q(a, 2a) is on this line, its coordinates will satisfy this equation.

  2a -4 = a -3

  a = 1 . . . . . . . . add 4-a to both sides

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The line with slope 2 on the attached diagram is the locus of all possible points Q. The one of interest is the one that is on the perpendicular through R.

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Additional comment

In order for a line to be an altitude to side QR, it must go through point P. The given line (y=-x+6) does not go through point P. We assume that the line y=-x+5 is the one that is intended.

225x²+4=0

225x²=0-4

225x²=-4

x²=-4/225

x=±√-4/225