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Coordinate Geometry

Short Answer Type

211. Find (he ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). Also find the point of intersection.

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212. Find the ratio in which the line segment joining the points (6, 4) and (1, - 7) is divided by x-axis. Also find the point of intersection.

2655 Views

213. If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram taken in order, find the value of p.

298 Views

214. Prove (hat the points (-2, -1), (1, 0), (4, 3) and (1, 2) are the vertices of a parallelogram is it a rectangle?

236 Views

215. The three vertices of a rhombus, taken in order are (2, -1), (3, 4) and (-2, 3). Find the fourth vertex, is it a square?

1151 Views

216. The mid-points of the sides of a triangle are (3, 4), (4, 1) and (2, 0). Find the coordinates of the vertices of the triangle.

89 Views

217. Find the lengths of the medians of the triangle whose vertices are (1, -1), (0, 4) and (-5, 3).

Let the given points of a triangle be A(1, -1), B(0, 4), C(-5, 3).
Let D, E, F be the mid-points of the sides BC, CA and AB respectively. Then,
The co-ordinates of D are :

$equals straight D open square brackets fraction numerator 0 plus 5 over denominator 2 end fraction comma space fraction numerator 4 plus 3 over denominator 2 end fraction close square brackets equals straight D open square brackets fraction numerator negative 5 over denominator 2 end fraction comma space 7 over 2 close square brackets$

the co-ordinates of E are

$equals straight E open square brackets fraction numerator negative 5 plus 1 over denominator 2 end fraction comma space fraction numerator 3 minus 1 over denominator 2 end fraction close square brackets equals space straight E space open square brackets fraction numerator negative 4 over denominator 2 end fraction comma space 2 over 2 close square brackets equals space straight E space left square bracket negative 2 comma space 1 right square bracket$

the co-ordinates of F are

$equals straight E space open square brackets fraction numerator 0 plus 1 over denominator 2 end fraction comma space fraction numerator 3 minus 1 over denominator 2 end fraction close square brackets equals straight E space open square brackets fraction numerator negative 4 over denominator 2 end fraction comma space 2 over 2 close square brackets equals straight E space left square bracket negative 2 comma space 1 right square bracket$

the co-ordinates of F are

$equals space straight F open square brackets fraction numerator 0 plus 1 over denominator 2 end fraction comma space fraction numerator 4 minus 1 over denominator 2 end fraction close square brackets equals space straight F space open square brackets 1 third comma space 3 over 2 close square brackets Now comma space space space AD space equals space square root of open parentheses fraction numerator negative 5 over denominator 2 end fraction minus 1 close parentheses squared plus open parentheses 7 over 2 plus 1 close parentheses squared end root space space space space space space space space space space space space space space space space space equals space square root of open parentheses fraction numerator negative 5 minus 2 over denominator 2 end fraction close parentheses squared plus open parentheses fraction numerator 7 plus 2 over denominator 2 end fraction close parentheses squared end root space space space space space space space space space space space space space space space space space equals space square root of open parentheses fraction numerator negative 7 over denominator 2 end fraction close parentheses squared plus open parentheses 9 over 2 close parentheses squared end root space space space space space space space space space space space space space space space space space equals space square root of 49 over 4 plus 81 over 4 end root equals square root of fraction numerator 49 plus 81 over denominator 4 end fraction end root equals square root of 130 over 4 end root space space space space space space space space space space space space space space space space space equals space square root of 130 over 2 end root space space units.$

$and space space space space CF space equals space square root of open parentheses 1 half plus 5 close parentheses squared plus open parentheses 3 over 2 minus 3 close parentheses squared end root space space space space space space space space space space space space space space equals space square root of open parentheses fraction numerator 1 plus 10 over denominator 2 end fraction close parentheses squared plus open parentheses fraction numerator 3 minus 6 over denominator 2 end fraction close parentheses squared end root space space space space space space space space space space space space space space equals space space square root of open parentheses 11 over 2 close parentheses squared plus open parentheses fraction numerator negative 3 over denominator 2 end fraction close parentheses squared end root space space space space space space space space space space space space space space equals space space space square root of open parentheses 121 over 4 plus 9 over 4 close parentheses end root space space space space space space space space space space space space space space equals space square root of fraction numerator 121 plus 9 over denominator 4 end fraction end root equals square root of 130 over 4 end root space space space space space space space space space space space space space space equals space fraction numerator square root of 130 over denominator 2 end fraction space units.$
Hence, the lengths of medians are

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