Advertisements
Advertisements
Question
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.
Advertisements
Solution
Let the given points be A(4, 4), B(3, 5) and C(–1, –1), then
slope of AB = `("y"_2 - "y"_1)/("x"_2 - "x"_1)`
= `(5 - 4)/(3 - 4)`
= `1/(-1)`
= −1 = m1
Slope of BC = `(-1 - 5)/(3 - 4)`
= `(-6)/(-4)`
= `3/2`
Slope of CA = `(4 + 1)/(4 + 1)`
= `5/5`
= 1 = m2
Slope of AB × Slope of CA = m1 × m2
= −1 × 1
= −1
Hence, AB ⊥ CA
= A, B, C are the vertices of a right triangle.
APPEARS IN
RELATED QUESTIONS
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.
If three point (h, 0), (a, b) and (0, k) lie on a line, show that `q/h + b/k = 1`
Find the values of k for which the line (k–3) x – (4 – k2) y + k2 –7k + 6 = 0 is
- Parallel to the x-axis,
- Parallel to the y-axis,
- Passing through the origin.
Find the slope of the lines which make the following angle with the positive direction of x-axis: \[\frac{\pi}{3}\]
Find the slope of a line passing through the following point:
(−3, 2) and (1, 4)
Find the slope of a line passing through the following point:
(3, −5), and (1, 2)
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)
State whether the two lines in each of the following is parallel, perpendicular or neither.
Through (6, 3) and (1, 1); through (−2, 5) and (2, −5)
Show that the line joining (2, −5) and (−2, 5) is perpendicular to the line joining (6, 3) and (1, 1).
Without using Pythagoras theorem, show that the points A (0, 4), B (1, 2) and C (3, 3) are the vertices of a right angled triangle.
Prove that the points (−4, −1), (−2, −4), (4, 0) and (2, 3) are the vertices of a rectangle.
If three points A (h, 0), P (a, b) and B (0, k) lie on a line, show that: \[\frac{a}{h} + \frac{b}{k} = 1\].
Consider the following population and year graph:
Find the slope of the line AB and using it, find what will be the population in the year 2010.
Find the value of x for which the points (x, −1), (2, 1) and (4, 5) are collinear.
A quadrilateral has vertices (4, 1), (1, 7), (−6, 0) and (−1, −9). Show that the mid-points of the sides of this quadrilateral form a parallelogram.
Find the equation of the perpendicular to the line segment joining (4, 3) and (−1, 1) if it cuts off an intercept −3 from y-axis.
Find the equations of the altitudes of a ∆ ABC whose vertices are A (1, 4), B (−3, 2) and C (−5, −3).
If the image of the point (2, 1) with respect to a line mirror is (5, 2), find the equation of the mirror.
If θ is the angle which the straight line joining the points (x1, y1) and (x2, y2) subtends at the origin, prove that \[\tan \theta = \frac{x_2 y_1 - x_1 y_2}{x_1 x_2 + y_1 y_2}\text { and } \cos \theta = \frac{x_1 x_2 + y_1 y_2}{\sqrt{{x_1}^2 + {y_1}^2}\sqrt{{x_2}^2 + {y_2}^2}}\].
Prove that the straight lines (a + b) x + (a − b ) y = 2ab, (a − b) x + (a + b) y = 2ab and x + y = 0 form an isosceles triangle whose vertical angle is 2 tan−1 \[\left( \frac{a}{b} \right)\].
Show that the tangent of an angle between the lines \[\frac{x}{a} + \frac{y}{b} = 1 \text { and } \frac{x}{a} - \frac{y}{b} = 1\text { is } \frac{2ab}{a^2 - b^2}\].
The acute angle between the medians drawn from the acute angles of a right angled isosceles triangle is
The reflection of the point (4, −13) about the line 5x + y + 6 = 0 is
If x + y = k is normal to y2 = 12x, then k is ______.
Find the equation of the straight line passing through (1, 2) and perpendicular to the line x + y + 7 = 0.
If the line joining two points A(2, 0) and B(3, 1) is rotated about A in anticlock wise direction through an angle of 15°. Find the equation of the line in new position.
Find the equation to the straight line passing through the point of intersection of the lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 and perpendicular to the line 3x – 5y + 11 = 0.
The coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0 are ______.
Slope of a line which cuts off intercepts of equal lengths on the axes is ______.
If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral.
The points A(– 2, 1), B(0, 5), C(– 1, 2) are collinear.
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.
