Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

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.

146 Views

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).

2113 Views

218. Determine the ratio in which the line 3x + y - 9 = 0 divides the segment joining the points (1, 3) and (2, 7).

436 Views

219. 26. If C is a point lying on the line segment AB joining A(1, 1) and B(2, - 3) seen that 3AC = CB, then find the co-ordinates of C.

We have,

3AC = CB

rightwards double arrow space space AC over CB equals 1 third

Now, coordinates of C are

straight C space open square brackets fraction numerator straight m subscript 1 straight x subscript 2 plus straight m subscript 2 straight x subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction comma space fraction numerator straight m subscript 1 straight y subscript 2 plus straight m subscript 2 straight y subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets

Here, we have x₁ = 1, y₁ = 1;
x₂ ^{= 2, y₂ = -3}and m₁ = 1, m₂ = 3
Hence,

$straight C space open square brackets fraction numerator 1 left parenthesis 2 right parenthesis plus left parenthesis 3 right parenthesis left parenthesis 1 right parenthesis over denominator 1 plus 3 end fraction comma space fraction numerator 1 left parenthesis negative 3 right parenthesis plus 3 left parenthesis 1 right parenthesis over denominator 1 plus 3 end fraction close square brackets equals straight C space open square brackets fraction numerator 2 plus 3 over denominator 4 end fraction comma space fraction numerator negative 3 plus 3 over denominator 4 end fraction close square brackets space equals space straight C open square brackets 5 over 4 comma space 0 close square brackets$

187 Views

220. Prove that the mid-point of the hypotenuse of a right angled triangle is equidistant from its vertices.

2329 Views