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Re: On the xy-plane, the xy-coordinate pairs (-6,2) and (-14, -4 [#permalink]
provide solution plz
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Re: On the xy-plane, the xy-coordinate pairs (-6,2) and (-14, -4 [#permalink]
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Slope of line passing through (-6,2) and (-14, -4) is = (-4-2) / (-14+6) = 3/4
Slope of line passing through (-12,1) and (-3, -11) is = (-11-1) / (-3+12) = -4/3

Eq. of a line in XY plane:
y = mx + b [m is the slope and (x,y) is a point on that line]

Y interceptor of line passing through (-6,2) and (-14, -4) is:

2 = (3 / 4) * -6 + b
b = 26 / 4

Y interceptor of line passing through (-12,1) and (-3, -11) is:
1 = (-4 / 3) * -12 + b
b = - 45 / 3

What is the unit length of the longest side of a triangle formed by the y-axis and these two lines is = 26 / 4 + 45 / 3 = 21.5
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On the xy-plane, the xy-coordinate pairs (-6,2) and (-14, -4 [#permalink]
2
GeminiHeat wrote:
On the xy-plane, the xy-coordinate pairs (-6,2) and (-14, -4) define one line, and the xy-coordinate pairs (-12,1) and (-3, -11) define another line. What is the unit length of the longest side of a triangle formed by the y-axis and these two lines?

(A) 15
(B) 17.5
(C) 19
(D) 21.5
(E) 23


Equation of line passing through (-6, 2) and (-14, -4);

\(y = \frac{(-4-2)}{(-14+6)}x + c\)

\(y = \frac{3}{4}x + c\)

y-intercept (c) = \(y - \frac{3}{4}x\)

\(c = 2 - \frac{3}{4}(-6) = \frac{13}{2}\)

Equation of line passing through (-12, 1) and (-3, -11);

\(y = \frac{(-11-1)}{(-3+12)}x + c\)

\(y = \frac{-4}{3}x + c\)

y-intercept (c) = \(y - \frac{-4}{3}x\)

\(c = 1 - \frac{-4}{3}(-12) = -15\)

Now, the distance between the y-intercepts = \(|\frac{13}{2}| + |-15| = 21.5\)

Hence, option D
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On the xy-plane, the xy-coordinate pairs (-6,2) and (-14, -4 [#permalink]
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