Advertisements
Advertisements
Question
Find the equations of the altitudes of a ∆ ABC whose vertices are A (1, 4), B (−3, 2) and C (−5, −3).
Advertisements
Solution
The vertices of ∆ABC are A (1, 4), B (−3, 2) and C (−5, −3).
Slope of AB = \[\frac{2 - 4}{- 3 - 1} = \frac{1}{2}\]
Slope of BC = \[\frac{- 3 - 2}{- 5 + 3} = \frac{5}{2}\]
Slope of CA = \[\frac{4 + 3}{1 + 5} = \frac{7}{6}\]
Thus, we have:
Slope of CF = \[- 2\]
Slope of AD = \[- \frac{2}{5}\]
Slope of BE = \[- \frac{6}{7}\]
Hence,
\[\text { Equation of CF is } : \]
\[y + 3 = - 2\left( x + 5 \right)\]
\[ \Rightarrow 2x + y + 13 = 0\]
\[\text { Equation of AD is } : \]
\[ y - 4 = - \frac{2}{5}\left( x - 1 \right) \]
\[ \Rightarrow 2x + 5y - 22 = 0\]
\[\text { Equation of BE is : } \]
\[ y - 2 = - \frac{6}{7}\left( x + 3 \right)\]
\[ \Rightarrow 6x + 7y + 4 = 0\]
APPEARS IN
RELATED QUESTIONS
Draw a quadrilateral in the Cartesian plane, whose vertices are (–4, 5), (0, 7), (5, –5) and (–4, –2). Also, find its area.
The base of an equilateral triangle with side 2a lies along they y-axis such that the mid point of the base is at the origin. Find vertices of the triangle.
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Without using distance formula, show that points (–2, –1), (4, 0), (3, 3) and (–3, 2) are vertices of a parallelogram.
Find the equation of a line drawn perpendicular to the line `x/4 + y/6 = 1`through the point, where it meets the y-axis.
Find the slope of the lines which make the following angle with the positive direction of x-axis:
\[\frac{2\pi}{3}\]
State whether the two lines in each of the following are parallel, perpendicular or neither.
Through (5, 6) and (2, 3); through (9, −2) and (6, −5)
Using the method of slope, show that the following points are collinear A (16, − 18), B (3, −6), C (−10, 6) .
What is the value of y so that the line through (3, y) and (2, 7) is parallel to the line through (−1, 4) and (0, 6)?
Prove that the points (−4, −1), (−2, −4), (4, 0) and (2, 3) are the vertices of a rectangle.
Line through the points (−2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x.
Find the value of x for which the points (x, −1), (2, 1) and (4, 5) are collinear.
Find the coordinates of the orthocentre of the triangle whose vertices are (−1, 3), (2, −1) and (0, 0).
Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Find the angles between the following pair of straight lines:
(m2 − mn) y = (mn + n2) x + n3 and (mn + m2) y = (mn − n2) x + m3.
Find the tangent of the angle between the lines which have intercepts 3, 4 and 1, 8 on the axes respectively.
Show that the line a2x + ay + 1 = 0 is perpendicular to the line x − ay = 1 for all non-zero real values of a.
The coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y − 11 = 0 are
Find k, if the slope of one of the lines given by kx2 + 8xy + y2 = 0 exceeds the slope of the other by 6.
If the slopes of the lines given by the equation ax2 + 2hxy + by2 = 0 are in the ratio 5 : 3, then the ratio h2 : ab = ______.
The equation of a line passing through the point (7, - 4) and perpendicular to the line passing through the points (2, 3) and (1 , - 2 ) is ______.
The coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0 are ______.
Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, – 1).
Find the angle between the lines y = `(2 - sqrt(3)) (x + 5)` and y = `(2 + sqrt(3))(x - 7)`
Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x + 4y = 4 and the opposite vertex of the hypotenuse is (2, 2).
The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x is ______.
One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is ______.
The vertex of an equilateral triangle is (2, 3) and the equation of the opposite side is x + y = 2. Then the other two sides are y – 3 = `(2 +- sqrt(3)) (x - 2)`.
The line `x/a + y/b` = 1 moves in such a way that `1/a^2 + 1/b^2 = 1/c^2`, where c is a constant. The locus of the foot of the perpendicular from the origin on the given line is x2 + y2 = c2.
The equation of the line through the intersection of the lines 2x – 3y = 0 and 4x – 5y = 2 and
| Column C1 | Column C2 |
| (a) Through the point (2, 1) is | (i) 2x – y = 4 |
| (b) Perpendicular to the line (ii) x + y – 5 = 0 x + 2y + 1 = 0 is |
(ii) x + y – 5 = 0 |
| (c) Parallel to the line (iii) x – y –1 = 0 3x – 4y + 5 = 0 is |
(iii) x – y –1 = 0 |
| (d) Equally inclined to the axes is | (iv) 3x – 4y – 1 = 0 |
The three straight lines ax + by = c, bx + cy = a and cx + ay = b are collinear, if ______.
