Slide 1 :
Slide 2 : Problem 2
The line segment joining (2, 3), (5, -1) is trisected.
Find the points of trisection. A(2, 3), B(5, -1)
AP:PB = 1:2 A(2,3) B(5,-1) P(x,y) Q(s,t) AQ:QB = 2:1
Proceed as above.
Slide 3 : Problem 3
Show that the figure formed by joining the midpoints of the sides of rectangle is a rhombus.
SlopeAB =-b/a = SlopeDC
SlopeBC = b/a = SlopeAD
ABCD is a parallelogram.
Hence the conclusion. O A(-a,0) C(a,0) D(0,b) B(0,-b)
Slide 4 : Problem 4
Prove that the points (a, 0), (0, b), (1, 1) are collinear if 1/a +1/b = 1
A(a, 0), B(0, b), C(1, 1)
SlopeAB=SlopeAC
-b/a=1/(1-a)
On simplification the result follows.
Slide 5 : Problem 5
A(-2, 7), B(1, 0), C(4, 3), D(1, 2) are the vertices of a quadrilateral ABCD.
Show that
SlopeAB = SlopeCD, SlopeBC = SlopeAD.
SlopeAB = -7/3, SlopeCD = 1/3 ? Wrong Question. Problem 6
Find the slope of a line parallel to the line joining the points (-4, 1), (2, 3)
Slope = 2/6=1/3
Slide 6 : Problem 7
Show that the line joining the points (2, -3),(-4, 1) is parallel to line joining the points (7, -1), (0, 3) and perpendicular to line joining the points (4, 5), (0, -2) respectively.
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A(2, -3), B(-4, 1)
C(7, -1), D(0, 3)
P(4, 5), Q(0, -2)
Hint: m1= slope AB,
m2 = slope CD, m3 = slope PQ
Show m1 = m2, m1.m3 = -1
Slide 7 : Problem 8
Find the locus of a point equidistant from the points (4, 2) and the x-axis.
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A(4, 2)
Let P(x, y) be the point.
AP^2 = y^2
(x-4)^2 +(y-2)^2 = y^2
Simplify.
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Slide 8 : Problem 9
A (2, 3), B (0, 2) are the coordinates of the two vertices of a triangle.
Find the locus of a point P such that the area of the triangle PAB = 3 sq. units
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Hint
P(x, y)
Area PAB = 3 P(x,y) A (2, 3) B(0, 2)
Slide 9 : Problem 10
Find the distance between the pair of points (-sin A, cos A) and (sin B, cos B).
---------------------------------- Problem 11
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5), (-1, -1) are the vertices of a right angled triangle.
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A(4, 4), B(3, 5), C(-1, -1), m1 = slopeAB, m2 = slopeBC, m3 = slopeCA. If m1m2 = -1 OR m2m3 = -1, m3m1 = -1 the conclusion follows.
Slide 10 : Problem 12
If A (2, -3), B (3, 5) are two vertices of a rectangle ABCD, find the slope of
BC, CD, DA.
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Slope AB =8, Slope BC = -1/8, slopeCD = 8, slope DA = -1/8
============================ Problem 13
Find the equation of the locus of a point equidistant from the points (2, 4) and y-axis.
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A(2, 4)
PA^2 = x^2
Slide 11 : Problem 14
If A (a, 0), B(-a, 0) are two fixed points, find the locus of a point P which moves so that 3|PA|= 2|PB|.
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9PA^2 = 4PB^2
9((x-a)^2+(y-0)^2)=4((x+a)^2+(y-0)^2)
Simplify Problem 15
Determine the equation of the line passing through the point (1, 2) which makes equal angles with the two axes.
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Slope = tan 45 = 1, y - y1 = m(x - x1)
Slide 12 : Problem 16
Find the equation of the line passing through (2, 3) and parallel to the line joining the points (2, -2), (6, 4).
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P(2, 3), A(2, -2), B(6, 4)
m=slopeAB
Y-y1 = m(x-x1)
======================== Problem 17
Find the equation of the medians of a triangle whose vertices are (2, 0), (0, 2), (6, 4).
A(2, 0), B(0, 2), C(6, 4) A(2, 0) B(0,2) C(6,4) D(3,3) E(4,2) F(1,1)
Slide 13 : Problem 18
Find the equation of a line such that the segment between the coordinate axes has its midpoint at the point (1, 3).
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P(1, 3)
1 = (a+0)/2, 3 = (b+0)/2
Equation is x/a + y/b = 1 P(1, 3) A(a,0) B(0,b) Problem 19
Find the equation of a line which passes through the point (3, -2) and cuts off positive intercepts on x and y axes in the ratio of 4:3.
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Intercepts 4a, 3a
Equation x/4a + y/3a = 1
Passes through (3, -2)
Find a
Slide 14 : Problem 20
If p is the length of the perpendicular segment from the origin, on the line whose intercepts on the axes are a, b, then show that 1/p^{2} = 1/a^{2} + 1/b^{2} .
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x/a + y/b = 1
bx + ay = ab
Distance from the origin
Slide 15 : Problem 21
Find the length of the line segment AB intercepted by the straight line 3x - 2y + 12 = 0 between the two axes.
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A(-4, 0), B(0,6)
Find AB
=================== Problem 22
Which of the lines 2x - y + 3 = 0, x - 4y - 7 = 0 is nearer from the origin?
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Slide 16 : Problem 23
What are the points on the axis of x whose perpendicular distance from the straight line x/4 + y/4 =1 is 3?
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x + y = 4
A(a,0)
Distance from A =
Slide 17 :