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.

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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.

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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 = (𝑚_1 𝑦_2 + 𝑚_2 𝑦_1)/(𝑚_1 + 𝑚_2 ) 0 = (𝑘 × 7 + 1 × (−6))/(𝑘 + 1) 0 = (7𝑘 − 6)/(𝑘 +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 = (𝑚1 𝑥2 + 𝑚2 𝑥1)/(𝑚1 + 𝑚2) = (6 × (−1) + 7 × (−4))/(6 + 7) = (−6 − 28)/13 = (−34)/13 Hence, the coordinate of point is P = (x, 0) = ((−𝟑𝟒)/𝟏𝟑,𝟎) 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 Δ ABC by two formulas Area of ΔABC = 1/2 × Base × Height Area of ΔABC = 1/2 × AB × h Also, we find Area Δ ABC by and then compare both areas Finding AB x1 = 4 , y1 = −2 x2 = 7 , y2 = 2 AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 7 −4)2+(2 −(−2))2) = √((3)2+(2+2)2) = √(32+(4)2) = √(9+16) = √25 = 5 Finding Area ΔABC 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 ΔABC = 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 Δ ABC by two formulas Area of ΔABC = 1/2 × Base × Height Area of ΔABC = 1/2 × AB × h Also, we find Area Δ ABC by coordinate geometry formula and then compare both areas Finding AB x1 = 4 , y1 = −2 x2 = 7 , y2 = 2 AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 7 −4)2+(2 −(−2))2) = √((3)2+(2+2)2) = √(32+(4)2) = √(9+16) = √25 = 5 Finding Area ΔABC 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 ΔABC = 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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