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Maths TNSB Mentoring
Class 10 Crash course
Coordinate geometry
Annual Revision V
3.
TBQ - Slope - point form of a staright line
Question:
2
m.
Find the equation of the line passing through the point \((2,-6)\) and having slope \(\frac{-3}{4}\).
Answer
:
The equation of a straight line when point and slope are given is computed using the formula:
\(y - y_1 = m(x - x_1)\)
\(y - y_1 = x - x_1\)
\(y - y_1 = m(x + x_1)\)
\(m(y - y_1) = x - x_1\)
The required equation of a straight line is:
\(3x + 4y + 18 = 0\)
\(3x - 4y - 18 = 0\)
\(3x - 4y + 18 = 0\)
\(3x + 4y - 18 = 0\)
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