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प्रश्न
Show that the points P (0, 5), Q (5, 10) and R (6, 3) are the vertices of an isosceles triangle.
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उत्तर
PQ = `sqrt((5 - 0)^2 + (10 - 5)^2`
= `sqrt(25+25)`
= `sqrt(50)`
= 5`sqrt(2)`
QR = `sqrt((6 - 5)^2 + (3 - 10)^2`
= `sqrt(1+49)`
= `sqrt(50)`
= 5`sqrt(2)`
RP = `sqrt((0 - 6)^2 + (5 - 3)^2`
= `sqrt(36+4)`
= `sqrt(40)`
= 2`sqrt(10)`
Since, PQ = QR, ΔPQR is an isosceles triangle.
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संबंधित प्रश्न
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Find the distance between the points `A((-8)/5, 2)` and `B(2/5, 2)`.
Find the distance between the following pair of point.
T(–3, 6), R(9, –10)
Find the distance between the following pairs of point in the coordinate plane :
(7 , -7) and (2 , 5)
Use distance formula to show that the points A(-1, 2), B(2, 5) and C(-5, -2) are collinear.
The point which divides the lines segment joining the points (7, -6) and (3, 4) in ratio 1 : 2 internally lies in the ______.
A circle has its centre at the origin and a point P(5, 0) lies on it. The point Q(6, 8) lies outside the circle.
The point P(–2, 4) lies on a circle of radius 6 and centre C(3, 5).
Find distance between points P(– 5, – 7) and Q(0, 3).
By distance formula,
PQ = `sqrt(square + (y_2 - y_1)^2`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(125)`
= `5sqrt(5)`
Find the distance between the points O(0, 0) and P(3, 4).
