in xyz pqr are mid point of xy,xz and yz bif PQ 6 pr 7.2find yz and xz
Answers
Answer࿐
Given,
3x+2y = -4 ----------(1)
2x+5y = 1 -----------(2)
¶ Find Point of Intersection of these 2 lines
Do 2×(1) - 3×(2)
6x+4y = -8 ----------(3)
6x+15y = 3 ---------(4)
-----------------------
-11y = -11
=> y = 1
Substitute in (1)
3x+2(1) = -4
=> 3x = -4-2
=> 3x = -6
=> x = -2
•°• Point of Intersection of the lines 3x+2y+4= 0 & 2x+5y-1= 0 is (-2,1)
¶ By using the Slope-Intercept form, find the form of equation of line passing through the point (-2,1)
y = mx + c
substitute x = -2 & y = 1
=> 1 = -2m + c
=> c = 1 + 2m
•°• The Required Equations of straight line is of form :
y = mx + 1 + 2m -----------(5)
¶ The perpendicular distance (or simply distance) 'd' of a point P(x1,y1) from Ax+By+C = 0 is given by
Given,
(x1,y1) = (-2,1) & d = 2
=> (4m+2)² = 2(m² + 1)
=> 16m² + 16m + 4 = 2m² + 2
=> 14m² + 16m + 2 = 0
=> 7m² + 8m + 1 = 0
Factorise the equation
=> 7m² + 7m + m + 1 = 0
=> 7m(m+1) + 1(m+1) = 0
=> (m+1)(7m+1) = 0
=> m = -1 and m = -1/7
Now Substitute m = -1 in (5)
=> y = (-1)x+1+2(-1)
=> y = -x + 1 - 2
=> y = -1 - x
(or)
=> -x - y -1 = 0
=> x + y + 1 = 0 ------------(6)
Substitute m = -1/7 in (5)
=> y = (-1/7)x + 1 + 2(-1/7)
=> y = -x/7 + (7-2)/7
=> y = (-x+5)/7
=> 7y = -x+5
or
=> -x - 7y + 5 = 0
=> x + 7y - 5 = 0 ----------(7)
•°• The Required equation of straight lines are :
x + y + 1 = 0 & x + 7y - 5 = 0
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Answer:
YZ= 12
XZ=14.4
Step-by-step explanation:
In triangle XYZ,
i) P and Q are midponits of XY and XZ
midpoint theorem ( the line passing through the midpoints of any 2 sides of a triangle is parallel and half of the third side)
by midpoint theorem,
PQ=1/2YZ
PQ*2/1=YZ
6*2/1=YZ
12=YZ
ii) P and R are midpoints of XY and YZ
by midpoint theorem,
PR=1/2XZ
PR*2/1=XZ
7.2*2=XZ
14.4 = XZ