If the points a x 2 b -3 4
WebSolution The given points are A (−2, 1), B ( a, b) and C (4, −1). Since the given points are collinear, the area of the triangle ABC is 0. ⇒ 1 2 [ x 1 ( y 2 - y 3) + x 2 ( y 3 - y 1) + x 3 ( y 1 - y 2)] = 0 Here, and x 1 = - 2, y 1 = 1, x 2 = a, y 2 = b and x 3 = 4, y 3 = - 1 ∴ 1 2 [ - 2 ( b + 1) + a ( - 1 - 1) + 4 ( 1 - b)] = 0 -2b-2-2a+4-4b=0 Web15 feb. 2024 · If A (x, 2), B (−3, −4), and C (7, −5) are collinear, then the value of x is (A) −63 (B) 63 (C) 60 (D) −60 This question is similar to Ex 7.3, 2 (i) - Chapter 7 Class 10 …
If the points a x 2 b -3 4
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Web29 mrt. 2024 · Transcript. Ex 7.2, 8 If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP = 3/7 AB & P lies on the line segment AB. Let the co−ordinates of point P be P (x, y) It is given that AP = 3/7 (AB) AP = 3/7 (AP + PB) 7AP = 3AP + 3PB 7AP − 3AP = 3PB 4AP = 3PB 𝐴𝑃/𝑃𝐵 = 3/4 Hence the point P ... WebMath Advanced Math 2. [-/3 Points] BBBASICSTAT8COREQ 5.R.013. Does it pay to ask for a raise? A national survey of heads of households showed the percentage of those …
WebThen, the three points are collinear if and only if the following $4\times 3$ matrix is column rank deficient: \begin{align} \left[ \begin{array}{ccc} p_1 & p_2 & p_3\\ 1 & 1 & 1\\ \end{array} \right]\in\mathbb{R}^{4\times 3} \end{align} Web26 jan. 2024 · If the points A (4,3) and B (x,5) are on the circle with Centre O (2,3) then find the value of x. Advertisement Expert-Verified Answer 2262 people found it helpful RahulCR7 point A (4,3) point B (X,5) CENTER O (2,3) DISTANCE BETWEEN A AND O distance between B and O is please mark it as brainliest answer. Find Math textbook solutions? …
Web12 nov. 2024 · Correct Answer - A. Here, (x1 = x, y1 = 2), (x2 = − 3, y2 = − 4)and(x3 = 7, y3 = − 5) ( x 1 = x, y 1 = 2), ( x 2 = - 3, y 2 = - 4) and ( x 3 = 7, y 3 = - 5) ∴ x1(y2 − y3) + … WebFind a relation between x and y, if the points Ax, y, B 5,7 and C 4,5 are collinear. [CBSE 2015] Login. Study Materials. NCERT Solutions. NCERT Solutions For Class 12. NCERT Solutions For Class 12 Physics; NCERT Solutions For Class 12 Chemistry; NCERT Solutions For Class 12 Biology;
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Web29 mrt. 2024 · Transcript. Ex 7.2 , 6 If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y. Let the points be A (1, 2), B (4, y), C (x, 6), D (3, 5) We know that diagonals of parallelogram bisect each other So, O is the mid−pint of AC & BD Finding mid−point of AC We have to find co−ordinates of O x− ... beast bagWebIf the points A (x,2),B (−3,−4),C (7,−5) are collinear, then find the value of x. Medium Solution Verified by Toppr A(X,2) B(−3,−4) C(7,−5) x=? slope AB = slope BC = slope AC … beast bank kendalWeb1 jan. 2024 · The given points are A (x, 2) , B (-3, -4) and C (7, -5).The points A, B and C are collinear if area of a triangle formed by joining these points is 0. so, area of Δ ABC = 0 ⇒12 [x1 (y2−y3) + x2 (y3−y1) + x3 (y1−y2)] = 0 ⇒12 [x (−4+5)−3 (−5−2) +7 (2+4)] = 0 ⇒12 [x − 3 (−7)+ 7×6] = 0 ⇒12 [x+21+42] = 0 ⇒12 [x+63] = 0 ⇒x+63 = 0 ⇒x = −63 beast bankWeb5 jun. 2024 · 5 Answers. Sorted by: 1. f(x) = ax3 − ax2 + bx + 4. Since f(4) = 51 , 51 = a(64 − 16) + 4b + 4 = 48a + 4b + 4 so 12a + b = 47 / 4. Since f(i) = 0 , 0 = a( − i + 1) + ib + 4 = i(b − a) + a + 4 so a + 4 = 0, a = − 4, b − a = 0, b = a = − 4. Therefore f(x) = − 4x3 + 4x2 − 4x + 4. But this does not satisfy f(4) = 51. did judy malinowski surviveWebHeight of a football A football is thrown by a quarterback from the 10-yard line and caught by the wide receiver on the 50-yard line. The footballs path on this interval can be modeled by the quadratic function f(x)=120x2+3x19, where x is the horizontal distance in yards from the goal line and f(x) is the height of the football in feet. did juicyfruitsnacks go to jailWeb12 nov. 2024 · Best answer Correct Answer - A Here, (x1 = x, y1 = 2), (x2 = − 3, y2 = − 4)and(x3 = 7, y3 = − 5) ( x 1 = x, y 1 = 2), ( x 2 = - 3, y 2 = - 4) and ( x 3 = 7, y 3 = - 5) ∴ x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2) = 0 ∴ x 1 ( y 2 - y 3) + x 2 ( y 3 - y 1) + x 3 ( y 1 - … did juice wrld take drugsWebSolution The required line passes through the points A (2, 0) = (x 1, y 1) and B (3, 4) = (x 2, y 2) say. Equation of the line in two point form is y - y 1 y 2 - y 1 = x - x 1 x 2 - x 1 ∴ the equation of the required line is y - 0 4 - 0 = x - 2 3 - 2 ∴ y 4 = x - 2 1 ∴ y = 4 (x – 2) ∴ y = 4x – 8 ∴ 4x – y – 8 = 0 beast au bsd manga