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प्रश्न
Find the equation of a straight line passing through the point P(−5, 2) and parallel to the line joining the points Q(3, −2) and R(−5, 4)
उत्तर
Slope of the line = `(y_2 - y_1)/(x_2 - x_1)`
Slope of the line QR = `(4 + 2)/(-5 - 3) = 6/(-8) = 3/(-4)`
⇒ `-3/4`
Slope of its parallel = `-3/4`
The given point is p(−5, 2)
Equation of the line is y – y1 = m(x – x1)
y – 2 = `-3/4(x + 5)`
4y – 8 = – 3x – 15
3x + 4y – 8 + 15 = 0
3x + 4y + 7 = 0
The equation of the line is 3x + 4y + 7 = 0
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