A line passes through the point (2, -1) and is perpendicular to the line \(y = \left(-\frac{2}{5}\right) x + 3\). What is its equation?
\(y=\left(\frac{5}{2}\right)x-6\)
The given line has slope \(m_1 = -\frac{2}{5}\).
For perpendicular lines \(m_1 m_2 = -1\), so \(m_2 = \frac{5}{2}\).
Using point-slope form through (2, −1): \(y + 1 = \frac{5}{2}(x - 2) \Rightarrow y = \frac{5}{2}x - 5 - 1\).
So \(y = \frac{5}{2}x - 6\).
Hence, the required equation is \(y = \left(\frac{5}{2}\right)x - 6\).
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Important Questions from Co-ordinate Geometry
In which ratio the point (-3, p) divides the line segment joining the points (-5, -4) and (-2, 3)?
In which quadrant both abscissa and ordinate are negative?
Find the slope of the line joining the points (3, -4) and (5, 2).
A1 (x1, y1), A2 (x2, y2), A3 (x3, y3) and A4 (x4, y4) are 4 points, p1 divides A1A2 in the ratio 1 ∶ 1, p2 divides P1A3 in the ratio 1 ∶ 2, P3 divides p2A4 in the ratio 1 ∶ 3. Then the coordinates of p3 are:
The polar coordinates of the vertices of a triangle are (0, 0), \(\left( {3,\frac{\pi }{2}} \right)\) and \(\left( {3,\frac{\pi }{6}} \right)\). Then, the triangle is:
