Advertisements
Advertisements
प्रश्न
Show that the points (−2, 3), (8, 3) and (6, 7) are the vertices of a right triangle ?
Advertisements
उत्तर
The given points are A (−2, 3), B (8, 3) and C (6, 7).
By, distance formula, AB `sqrt{{8-(-2)}^2+(3-3)^2`
`rArr AB^2=10^2+0`
`rArrBC^2=100`
`BC=sqrt((6-8)^2+(7-3)^2)`
`rArrBC^2=(-2)^2+4^2`
`rArrBC^2=4+16`
`rArrBC^2=20`
`BC=sqrt((-2-6)^2+(3-7)^2)`
`rArrCA^2=(-8)^2+(-4)^2`
`rArrCA^2=64+16`
`rArrCA^2=80`
It is observed that BC2 + CA2 = 20 + 80 = 100 = AB2
So, by the converse of Pythagoras Theorem, ΔABC is right angled at C.
APPEARS IN
संबंधित प्रश्न
Prove that the points (0, 0), (5, 5) and (-5, 5) are the vertices of a right isosceles triangle.
Find the coordinates of the midpoints of the line segment joining
P(-11,-8) and Q(8,-2)
Find the ratio in which the point (-1, y) lying on the line segment joining points A(-3, 10) and (6, -8) divides it. Also, find the value of y.
Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are A(2,1) B(4,3) and C(2,5)
A point whose abscissa and ordinate are 2 and −5 respectively, lies in
If (a,b) is the mid-point of the line segment joining the points A (10, - 6) , B (k,4) and a - 2b = 18 , find the value of k and the distance AB.
If three points (x1, y1) (x2, y2), (x3, y3) lie on the same line, prove that \[\frac{y_2 - y_3}{x_2 x_3} + \frac{y_3 - y_1}{x_3 x_1} + \frac{y_1 - y_2}{x_1 x_2} = 0\]
Two vertices of a triangle have coordinates (−8, 7) and (9, 4) . If the centroid of the triangle is at the origin, what are the coordinates of the third vertex?
The perpendicular distance of the point P(3, 4) from the y-axis is ______.
Points (1, –1) and (–1, 1) lie in the same quadrant.
