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Practice each question, then open the answer or worked solution.
Name the type of triangle formed by the points A (–5, 6), B (–4, –2) and C (7, 5).
The points are A (–5, 6), B (–4, –2) and C (7, 5)
Using distance formula,
d = √ ((x2 – x1)2 + (y2 – y1)2)
AB = √((-4+5)² + (-2-6)²)
= √1+64
=√65
BC=√((7+4)² + (5+2)²)
=√121 + 49
=√170
AC=√((7+5)² + (5-6)²)
=√144 + 1
=√145
Since all sides are of different length, ABC is a scalene triangle.
Find the points on the x–axis which are at a distance of 2√5 from the point (7, –4). How many such points are there?
Let coordinates of the point=(x, 0)(given that the point lies on x axis)
x1=7. y1=-4
x2=x. y2=0
Distance =√(x2-x1)2+ (y2-y1)2
According to the question,
2√5=√(x-7)2+ (0-(-4))2
Squaring L.H.S and R.H.S
20=x2+49-14x+16
20=x2+65-14x
0=x2-14x+45
0=x2-9x-5x+45
0=x(x-9)-5(x-9)
0=(x-9) (x-5)
x-9 =0. x-5= 0
x=9 or x=5
Therefore, coordinates of points…..(9,0)or(5,0)
What type of a quadrilateral do the points A (2, –2), B (7, 3), C (11, –1) and D (6, –6) taken in that order, form?
The points are A (2, –2), B (7, 3), C (11, –1) and D (6, –6)
Using distance formula,
d = √ ((x2 – x1)2 + (y2 – y1)2)
Finding diagonals AC and BD, we get,
The Quadrilateral formed is rectangle.
Find the value of a, if the distance between the points A (–3, –14) and B (a, –5) is 9 units.
Distance between two points (x1,y1) ( x2,y2) is :
d=√(x2-x1)²+(y2-y1)²
Distance between A (–3, –14) and B (a, –5) is :
=√[(a+3)²+(-5+14)²] =9
Squaring on L.H.S and R.H.S.
(a+3)²+81=81
(a+3)²=0
(a+3)(a+3)=0
a+3=0
a= -3
Find a point which is equidistant from the points A (–5, 4) and B (–1, 6)? How many such points are there?
Let the point be P
According to the question,
P is equidistant from A (–5, 4) and B (–1, 6)
Then the point P = ((x1+x2)/2, (y1+y2)/2)
= ((-5-1)/2, (6+4)/2)
= (-3 , 5 )
Find the coordinates of the point Q on the x–axis which lies on the perpendicular bisector of the line segment joining the points A (–5, –2) and B(4, –2). Name the type of triangle formed by the points Q, A and B.
Point Q is the midpoint of AB as the point P lies on the perpendicular bisector of AB.
By mid point formula:
(x1 + x2)/2 = (-5+4)/2
= -½
x = -½
Given that, P lies on x axis, so y=0
P(x,y)= (-½ , 0)
Therefore, it is an isosceles triangle
Find the value of m if the points (5, 1), (–2, –3) and (8, 2m) are collinear.
The points A(5, 1), B(–2, –3) and C(8, 2m) are collinear.
i.e., Area of ∆ABC = 0
½ [x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 )]=0
½ [5(-3 – 2m) + ( – 2)(2m – 1) + 8(1 – ( – 3))]=0
½ (-15 – 10m – 4m + 2 + 32) = 0
½ (-14m + 19) = 0
m = 19/14
If the point A (2, – 4) is equidistant from P (3, 8) and Q (–10, y), find the values of y. Also find distance PQ.
Find the area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9).
Given vertices are:
(x₁, y₁) = (-8, 4)
(x₂, y₂) = (-6, 6)
(x₃, y₃) = (-3, 9)
Area of triangle = (½) (x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂))
= (½) (-8(6 – 9) + -6(9 – 4) + -3(4 – 6))
= (½) (-8(-3) + -6(5) + -3(-2))
= (½) (24 – 30 + 6)
= (½) (30 – 30)
= 0 units.
In what ratio does the x–axis divide the line segment joining the points (– 4, – 6) and (–1, 7)? Find the coordinates of the point of division.
Let the ratio in which x-axis divides the line segment joining (–4, –6) and (–1, 7) = 1: k.
Then,
x-coordinate becomes (-1 – 4k) / (k + 1)
y-coordinate becomes (7 – 6k) / (k + 1)
Since P lies on x-axis, y coordinate = 0
(7 – 6k) / (k + 1) = 0
7 – 6k = 0
k = 6/7
Now, m1 = 6 and m2 = 7
By using section formula,
x = (m1x2 + m2x1)/(m1 + m2)
= (6(-1) + 7(-4))/(6+7)
= (-6-28)/13
= -34/13
So, now
y = (6(7) + 7(-6))/(6+7)
= (42-42)/13
= 0
Hence, the coordinates of P are (-34/13, 0)