que no.24 plzzz

que no.24 plzzz 212 20. 21. 22. 23. 25. 26. 27. 28. 29. 31. 33. 2. 3. 4. A Class Book of Mathematics - The area of a triangle is S sq. units. Two or its vertices are (2. l) and (3, Find the third 93, (x, y) if x + 3 y. The vertices ofa triangle ABC are S), (3. l) and (5, 7). Find the area of the triangle IHS911 joining the mid.points of the sides of &4BC. Show that the points 3). B (-1, -S) and C (2, —S) form a right angled triangle. If a ABCD be completed. find the coordinates ofD. The line segment joining the points (2, -2) and (4, 6) is extended at both ends uplo half of i length. Find the coordinates of the extreme points. If P divides the side BC of AOBC in the ratio m : n. prove that • BC2 + (m + n) • n • OB2 + m • OC2 = If two medians of a triangle are equal, using coordinate geometry of two dimensions. prove tk triangle is isosceles. Find the area of the quadrilateral formed by the points (l, 2), (=2. 3), (2. 4) and (3, 4) taken i order. Find the equation to the locus of a point whose distances from two fixed points 0) and (a, l) are in a constant ratio k. Find the equation to the locus of a point which is equidistant from the points (l, l) and (-1. -li Ills n Find the equation to the locus of a point whose distance from the X-axis is double its the Y-axis. rod AB of length a + b slides in the 1st quadrant such that its ends A and B remain in contxt • the axes. Find the equation to the locus ofa moving point P on AB such that AP : PB = b Given three points A (a, O), B O), C (c, 0); find the equation to the locus of the point p if PA2 + PB2 2PC2. 32. A O) and B (a, O) are two fixed points. Show that the locus ofa point P such that CHA 121 Sl directio: line ant and if called t! The equation x2 + y: O does not represent any locus — Give reasons. (0.0), (10. 0). (10, 5), (0, 5) (i) right-angled (1) Square (ii) equilateral (ii) rhombus ANSWERS (iii) isosceles (iii) parallelogram (iv) rectangle 14

Dear student,
OBC in the ratio m:n n×OB2+m×OC2=mnm+n×BC2+m+n×OP2use cosB = a2+b2-c22abinOPBcosB = OB2+mm+n2BC2-OP22OB×mm+nBCinOCBcosB = OB2+BC2-OC22OB×BCcosB = OB2+mm+n2BC2-OP22OB×mm+nBC=cosB = OB2+BC2-OC22OB×BCOB2+BC2-OC2mm+n = OB2+mm+n2BC2-OP2OB2+BC2-OC2= m+nmOB2+mm+nBC2-m+nmOP2 ....1in OCPuse cosC = a2+c2-b22accosC = OC2+nm+n2BC2-OP22OCnm+nBCin OCBcosC = OC2+BC2-OB22OC×BCcosC = OC2+nm+n2BC2-OP22OCnm+nBC=cosC = OC2+BC2-OB22OC×BCOC2+BC2-OB2nm+n =OC2+nm+n2BC2-OP2 OC2+BC2-OB2 =m+nnOC2+nm+nBC2-m+nnOP2   ..eq1eq1+eq22BC2 = m+nnOC2+nm+nBC2-m+nnOP2+m+nmOB2+mm+nBC2-m+nmOP22BC2 = m+nnOC2+BC2-m+nnOP2+m+nmOB2-m+nmOP2BC2 = m+nnOC2-m+nnOP2+m+nmOB2-m+nmOP2mnm+nBC2 = mOC2-mOP2+nOB2-nOP2mnm+nBC2 +m+nOP2= mOC2+nOB2

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