Advertisements
Advertisements
Question
If A(1, –1), B(0, 4), C(–5, 3) are vertices of a triangle then find the slope of each side.
Solution
A (1, –1), B (0, 4), C (–5, 3) form a triangle.
\[\text{We know that, slope of line}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\]
Slope of AB = \[\frac{4 - \left( - 1 \right)}{0 - 1} = \frac{5}{- 1} = - 5\]
Slope of BC = \[\frac{3 - 4}{- 5 - 0} = \frac{- 1}{- 5} = \frac{1}{5}\]
Slope of AC = \[\frac{3 - \left( - 1 \right)}{- 5 - 1} = \frac{4}{- 6} = \frac{- 2}{3}\]
∴ The slopes of the sides AB, BC and AC are -5, \[\frac{1}{5}\] and \[\frac{-2}3\] respectively.
APPEARS IN
RELATED QUESTIONS
Without using the distance formula, show that the points A(4, −2), B(−4, 4) and C(10, 6) are the vertices of a right-angled triangle.
(−2, 4), (4, 8), (10, 7) and (11, –5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.
Find x, if the slope of the line joining (x, 2) and (8, −11) is `−3/4`.
The points (K, 3), (2, −4) and (−K + 1, −2) are collinear. Find K.
Find the slope and the inclination of the line AB if : A = `(-1, 2sqrt(3))` and B = `(-2, sqrt(3))`
Find the slope of the line which is parallel to `x/2 - y/3 -1 = 0 `
Find the value of p if the lines, whose equations are 2x – y + 5 = 0 and px + 3y = 4 are perpendicular to each other.
Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other.
The ordinate of a point lying on the line joining the points (6, 4) and (7, –5) is –23. Find the coordinates of that point.
Fill in the blank using correct alternative.
A line makes an angle of 30° with the positive direction of X– axis. So the slope of the line is ______.
Find the slope of the diagonals of a quadrilateral with vertices A(1, 7), B(6, 3), C(0, –3) and D(–3, 3).
Find the slope of a line, correct of two decimals, whose inclination is 50°
Find the slope and the y-intercept of the following line 5x - 2y = 6
Find the slope and the y-intercept of the following line x - 2 = `(5 - 3"y")/2`
Find the slope of the line passing through the points A(4,7) and B(2,3).
With out Pythagoras theorem, show that A(4, 4), B(3, 5) and C(-1, -1) are the vertices of a right angled.
Determine whether the following points are collinear. A(–1, –1), B(0, 1), C(1, 3)
Given: Points A(–1, –1), B(0, 1) and C(1, 3)
Slope of line AB = `(square - square)/(square - square) = square/square` = 2
Slope of line BC = `(square - square)/(square - square) = square/square` = 2
Slope of line AB = Slope of line BC and B is the common point.
∴ Points A, B and C are collinear.
