In what ratio does the x-axis divide the line segment joining the points (–4, –6) and (–1, 7)? Find the co-ordinates of the point of division.

OR

The points A(4, –2), B(7, 2), C(0, 9) and D(–3, 5) form a parallelogram. Find the length of the altitude of the parallelogram on the base AB.

This is a question of CBSE Sample Paper - Class 10 - 2017/18.

You can download the question paper here  https://www.teachoo.com/cbse/sample-papers/

In what ratio does x-axis divide line segment | The points A (4, -2)

Question 16 - CBSE Class 10 Sample Paper for 2018 Boards - Part 2
Question 16 - CBSE Class 10 Sample Paper for 2018 Boards - Part 3 Question 16 - CBSE Class 10 Sample Paper for 2018 Boards - Part 4 Question 16 - CBSE Class 10 Sample Paper for 2018 Boards - Part 5 Question 16 - CBSE Class 10 Sample Paper for 2018 Boards - Part 6 Question 16 - CBSE Class 10 Sample Paper for 2018 Boards - Part 7 Question 16 - CBSE Class 10 Sample Paper for 2018 Boards - Part 8 Question 16 - CBSE Class 10 Sample Paper for 2018 Boards - Part 9

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Transcript

Question 16 In what ratio does the x-axis divide the line segment joining the points ( 4, 6) and ( 1, 7)? Find the co-ordinates of the point of division. Let A( 4, 6) and B ( 1, 7) Let P be the point on x-axis which divides AB Since Point P is on x axis, Hence its y coordinate is 0. So, it is of the form P(x, 0) Now, we have to find ratio Let ratio be k : 1 Hence, m1 = k , m2 = 1 x1 = 4 , y1 = 6 x2 = 1 , y2 = 7 Also, x = x , y = 0 Using section formula y = ( _1 _2 + _2 _1)/( _1 + _2 ) 0 = ( 7 + 1 ( 6))/( + 1) 0 = (7 6)/( +1) 0(k + 1)= 7k 6 0 = 7k 6 7k 6 = 0 7k = 6 k = 6/7 Hence, Ratio = k : 1 Ratio = 6/7 : 1 i.e. 6 : 7 Now, we need to find x also x = ( 1 2 + 2 1)/( 1 + 2) = (6 ( 1) + 7 ( 4))/(6 + 7) = ( 6 28)/13 = ( 34)/13 Hence, the coordinate of point is P = (x, 0) = (( )/ , ) Question 16 The points A(4, 2), B(7, 2), C(0, 9) and D( 3, 5) form a parallelogram. Find the length of the altitude of the parallelogram on the base AB. Given ABCD forms a parallelogram We need to find length of altitude on base AB Let us draw an altitude CE from point C to AB Let CE = h We need to find h To find h, we find Area of ABC by two formulas Area of ABC = 1/2 Base Height Area of ABC = 1/2 AB h Also, we find Area ABC by and then compare both areas Finding AB x1 = 4 , y1 = 2 x2 = 7 , y2 = 2 AB = (( 2 1)2+( 2 1)2) = (( 7 4)2+(2 ( 2))2) = ((3)2+(2+2)2) = (32+(4)2) = (9+16) = 25 = 5 Finding Area ABC Area of triangle ABC = 1/2 [ x1(y2 y3) + x2(y3 y1) + x3(y1 y2) ] Here x1 = 4 , y1 = 2 x2 = 7 , y2 = 2 x3 = 0 , y3 = 9 Putting values Area of triangle ABC = 1/2 [ 4(2 9) + 7(9 ( 2)) + 0( 2 2) ] = 1/2 [ 4( 7) + 7(9 + 2) + 0( 4) ] = 1/2 [ 4( 7) + 7(11) + 0( 4) ] = 1/2 [ 28 + 77] = 49/2 Now, From (1) Area of ABC = 1/2 AB h Putting values of Area and AB 49/2 = 1/2 5 h 49/2 2/5 = h 49/5 = h h = 49/5 h = 9.8 units Question 16 The points A(4, 2), B(7, 2), C(0, 9) and D( 3, 5) form a parallelogram. Find the length of the altitude of the parallelogram on the base AB. Given ABCD forms a parallelogram We need to find length of altitude on base AB Let us draw an altitude CE from point C to AB Let CE = h We need to find h To find h, we find Area of ABC by two formulas Area of ABC = 1/2 Base Height Area of ABC = 1/2 AB h Also, we find Area ABC by coordinate geometry formula and then compare both areas Finding AB x1 = 4 , y1 = 2 x2 = 7 , y2 = 2 AB = (( 2 1)2+( 2 1)2) = (( 7 4)2+(2 ( 2))2) = ((3)2+(2+2)2) = (32+(4)2) = (9+16) = 25 = 5 Finding Area ABC Area of triangle ABC = 1/2 [ x1(y2 y3) + x2(y3 y1) + x3(y1 y2) ] Here x1 = 4 , y1 = 2 x2 = 7 , y2 = 2 x3 = 0 , y3 = 9 Putting values Area of triangle ABC = 1/2 [ 4(2 9) + 7(9 ( 2)) + 0( 2 2) ] = 1/2 [ 4( 7) + 7(9 + 2) + 0( 4) ] = 1/2 [ 4( 7) + 7(11) + 0( 4) ] = 1/2 [ 28 + 77] = 49/2 Now, From (1) Area of ABC = 1/2 AB h Putting values of Area and AB 49/2 = 1/2 5 h 49/2 2/5 = h 49/5 = h h = 49/5 h = 9.8 units

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.