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1999-30: Find the tangent of the acute angle between the lines 2x+y=3 and 3x-2y=5
Solution
From line 2x+y=3
y=-2x+3
{ m }_{ 1 }=-2
From line 3x-2y=5
2y=3x-5y
{ m }_{ 2 }=3
tan\theta =\frac { { m }_{ 2 }-{ m }_{ 1 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \\ \\ \\ tan\theta =\frac { 3+2 }{ 1+(-6) } \\ \\ \\ tan\theta =\frac { 5 }{ 1-6 } \\ \\ \\ tan\theta =\frac { 5 }{ -5 } \\ \\ \\ tan\theta =-1
2001-24: Find the value of p if the line joining (p,40 and (6,-2) is perpendicular to the line joining (2,p) and (1,3)
solution
Let the gradient of the line joining (p,4) and (6,-2) be { m }_{ 1 }
and the gradient of the line joining (2,p) and (1,3) be { m }_{ 2 }
{ m }_{ 1 }{ m }_{ 2 }=-1 for perpendiculiarity
{ m }_{ 1 }=\frac { -2-4 }{ 6-p } =\frac { -6 }{ 6-p } \\ { m }_{ 2 }=\frac { 3-p }{ -1-2 } =\frac { 3-p }{ -3 } \\ \frac { -6 }{ 6-p } \times \frac { 3-p }{ -3 } =-1\\ \frac { -18+6p }{ -18+3p } =-1\\ -18+6p=18+3p\\ 6p-3p=18+18\\ 9p=36\\ p=4
2001-30: P(-6,1) and Q (6,6) are the two ends of the diameter of a given circle. Calculate the radius
Solution
Distance between the points will give the diameter. Half of it will give us the radius
Distance(r)=\sqrt { { (6+6) }^{ 2 }+{ (6-1) }^{ 2 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad r=\sqrt { { 12 }^{ 2 }+{ 5 }^{ 2 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad r=\sqrt { 144+25 } \\ \quad \quad \quad \quad \quad \quad \quad \quad r=\sqrt { 169 } \\ \quad \quad \quad \quad \quad \quad \quad \quad r=13
1994-36: The equation of the line in the graph is (check for the diagram on your jamb past question)
Solution
The coordinates are (0,4) and (3,0)
\frac { y-{ y }_{ 1 } }{ x-{ x }_{ 1 } } =\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \\ \frac { y-4 }{ x-0 } =\frac { 0-4 }{ 3-0 } \\ \frac { y-4 }{ x } =\frac { -4 }{ 3 } \\ 3y-12=-4x\\ 3y=-4x+12\\ 3y+4x=12
1997-33: The angle between the positive horizontal axis and a given line is { 135 }^{ \circ }. Find the equation of the line if it passes through the point (2,3)
Solution
gradient(m)=tan\theta \\ \quad \quad \quad \quad \quad \quad \quad m=tan{ 135 }^{ \circ }\\ \quad \quad \quad \quad \quad \quad \quad m=-tan{ 45 }^{ \circ }\\ \quad \quad \quad \quad \quad \quad \quad m=-1\\ m=\cfrac { y-{ y }_{ 1 } }{ x-{ x }_{ 1 } } \\ -1=\frac { y-3 }{ x-2 } \\ y-3=-x+2\\ y+x=5
1997-34: Find the distance between the point Q(4,3) and the point common to the lines 2x-y=4 and x+y=2
Solution
2x-y=4……….eq(1)
x+y=2………..eq(2)
adding both equations give
3x=6
x=2
substitute x=2 into eq(2)
2+y=2
y=0
points common to the lines = (2,0)
Distance from point Q(4,3) is given as
\sqrt { { (4-2) }^{ 2 }+{ (3-0) }^{ 2 } } =\sqrt { 4+9 } \\ Distance=\sqrt { 13 }
1998-32: If the distance between the points (x,3) and (-x,2) is 5. Find x
Solution
\sqrt { { (-x-x) }^{ 2 }+{ (2-3 })^{ 2 } } =5\\ { (-2x })^{ 2 }+{ -1 }^{ 2 }={ 5 }^{ 2 }\\ 4{ x }^{ 2 }+1=25\\ 4{ x }^{ 2 }=24\\ 4{ x }^{ 2 }-24=0\\ 4({ x }^{ 2 }-6)=0\\ { x }^{ 2 }-6=0\\ { x }^{ 2 }=6\\ x=\pm \sqrt { 6 } \\ since\quad x\quad cannot\quad be\quad negative,\\ x=\sqrt { 6 }
1982-43: If the function y=5x is graphed, what would be the intercept on the y-axis?
Solution
y=mx+c…………eq(1)
where m is the gradient and c is the intercept on the y-axis
y=5x………………eq(2)
comparing both equations,
m=5 and c=0
\therefore intercept on the y-axis is 0
1984-11: A variable point P(x,y) traces a graph in a two-dimensional plane. (0,-3) is one position of p. If x increases by 1unit, y increases by 4units. The equation of the graph is?
Solution
(x,y)=(0,-3)
{ x }_{ 2 }-{ x }_{ 1 }=1\\ { y }_{ 2 }-{ y }_{ 1 }=4\\ Equation\quad of\quad the\quad graph\quad is\quad given\quad by\\ \frac { x-{ x }_{ 1 } }{ y-{ y }_{ 1 } } =\frac { { x }_{ 2 }-{ x }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } \\ \frac { x-0 }{ y+3 } =\frac { 1 }{ 4 } \\ 4x=y+3\\ y=4x-3
1986-30: At what points does the straight-line y=2x+1 intersect the curve y=2{ x }^{ 2 }+5x-1
Solution
2{ x }^{ 2 }+5x-1=2x+1\\ 2{ x }^{ 2 }+3x-2=0\\ 2{ x }^{ 2 }+4x-x-2=0\\ (2{ x }^{ 2 }+4x)-(x+2)=0\\ 2x(x+2)-1(x+2)=0\\ (x+2)(2x-1)=0\\ x+2=0\quad or\quad 2x-1=0\\ x=-2\quad or\quad \frac { 1 }{ 2 } \\ Using\quad one\quad of\quad the\quad equations,\\ if\quad x=-2\\ y=2(-2)+1\\ y=-3\\ point(-2,-3)\\ if\quad x=\frac { 1 }{ 2 } \\ y=2(\frac { 1 }{ 2 } )+1\\ y=2\\ point(\frac { 1 }{ 2 } ,2)\\ \therefore \quad the\quad points\quad are\quad (-2,-3)\quad and\quad (\frac { 1 }{ 2 } ,2)
1991-26: Find the gradient of the line passing through the points (-2,0) and (0,-4)
Solution
\frac { y-{ y }_{ 1 } }{ x-{ x }_{ 1 } } =\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \\ \frac { y-0 }{ x+2 } =\frac { -4-0 }{ 0+2 } \\ \frac { y }{ x+2 } =\frac { -4 }{ 2 } \\ \frac { y }{ x+2 } =-2\\ y=-2x-4\\ comparing\quad the\quad equation\quad to\quad y=mx+c\\ m=-2\\ \therefore gradient=-2\\
1992-32: What is the perpendicular distance of a point (2,3) from the line 2x-4y+3=0?
Solution
2x-4y+3=0\\ 4y=2x+3\\ y=\frac { 2x+3 }{ 4 } \\ y=\frac { 1 }{ 2 } x+\frac { 3 }{ 4 } \\ { m }_{ 1 }=\frac { 1 }{ 2 } \\ For\quad perpendiculiarity,\quad { m }_{ 1 }{ m }_{ 2 }=-1\\ \therefore { m }_{ 2 }=-2
1992-33: Find the equation of the line through (5,7) parallel to the line 7x+5y=12
Solution
7x+5y=12\\ 5y=12-7x\\ y=\frac { 12-7x }{ 5 } \\ y=\frac { -7 }{ 5 } x+\frac { 12 }{ 5 } \\ { m }_{ 2 }=\frac { -7 }{ 5 } \\ For\quad parallel\quad lines,\quad { m }_{ 1 }={ m }_{ 2 }\\ { m }_{ 1 }=\frac { y-{ y }_{ 1 } }{ x-{ x }_{ 1 } } \\ \frac { -7 }{ 5 } =\frac { y-7 }{ x-5 } \\ -7x+35=5y-35\\ 5y+7x=70
1994-35: If M(4,q) is the midpoint of the line joining L(p,-2) and N(q,p), find the values of p and q
Solution
x:\frac { p+q }{ 2 } =4\quad \quad \quad \quad \quad \quad \quad \quad y:\frac { -2+p }{ 2 } =q\\ p+q=8\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -2+p=2q\\ \quad \quad \quad \quad \quad \quad p+q=8.........eq(1)\\ \quad \quad \quad \quad \quad \quad p-2q=2........eq(2)\\ subtracting\quad eq(2)\quad from\quad eq(1)\quad gives\\ 3q=6\\ q=2\\ substituting\quad q=2\quad into\quad eq(1)\quad gives\\ p+2=8\\ p=6\\ \therefore p=6,\quad q=2
2002-49:Find the value of \alpha if the line 2y-\alpha x+k=0 is perpendicular to y+\frac { 1 }{ 4 } x-7=0
Solution
2y-\alpha x+k=0\\ 2y=\alpha x-k\\ y=\frac { \alpha }{ 2 } x-k\\ { m }_{ 1 }=\frac { \alpha }{ 2 } \\ y+\frac { 1 }{ 4 } x-7=0\\ y=\frac { -1 }{ 4 } x+7\\ { m }_{ 2 }=\frac { -1 }{ 4 } \\ { m }_{ 1 }{ m }_{ 2 }=-1\quad for\quad perpendiculiarity\\ \frac { \alpha }{ 2 } \times \frac { -1 }{ 4 } =-1\\ \frac { \alpha }{ 2 } =4\\ \alpha =8
2003-33: A triangle has vertices P(-1,6), Q(-3,-4) and R(1,-4). Find midpoints of PQ and QR respectively.
Solutions
For\quad PQ,\quad midpoint=(\frac { -1-3 }{ 2 } ,\frac { 6-4 }{ 2 } )\Rightarrow (2,1)\\ For\quad QR,\quad midpoint=(\frac { 1-3 }{ 2 } ,\frac { -4-4 }{ 2 } )\Rightarrow (-1,-4)
2004-31: Find the value of { \alpha }^{ 2 }+{ \beta }^{ 2 } if { \alpha }+{ \beta }=2 and the distance between the points (1,\alpha ) and (\beta, 1) is 3 units
Solution
distance\quad between\quad the\quad points=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } \\ 3=\sqrt { { (\beta -1) }^{ 2 }+{ (1-\alpha ) }^{ 2 } } \\ 3={ \beta }^{ 2 }-2\beta +1+1-2\alpha +{ \alpha }^{ 2 }\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }-2(\alpha +\beta )=3-2\\ recall\quad \alpha +\beta =2\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }-2(2)=1\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }-4=1\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }=5
2004-32: Find the midpoint of the line joining P(-3,5) and Q(5,-3)
Solution
midpoint=(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } )\\ =(\frac { -3+5 }{ 2 } ,\frac { 5-3 }{ 2 } )\\ =(1,1)
2005-32: Find the equation of the perpendicular at point (4,3) to the line y+2x=5
Solution
y=2x+5\\ y=-2x+5\\ { m }_{ 1 }=-2\\ { m }_{ 1 }{ m }_{ 2 }=-1\quad for\quad perpendiculiarity\\ -2{ m }_{ 2 }=-1\\ { m }_{ 2 }=\frac { 1 }{ 2 } \\ { m }_{ 2 }=\frac { y-3 }{ x-4 } \\ \frac { 1 }{ 2 } =\frac { y-3 }{ x-4 } \\ x-4=2(y-3)\\ x-4=2y-6\\ 2y-x=2
2005-33: Find the coordinate of the midpoint of the line joining (3,-4) and (-1,10)
Solution
M=(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } )\\ M=(\frac { 3-1 }{ 2 } ,\frac { -4+10 }{ 2 } )\\ M=(\frac { 2 }{ 2 } ,\frac { 6 }{ 2 } )\\ M=(1,3)
2006-8: PQ and RS are two parallel lines. If the coordinates of P,Q,R,S are (1,q), (3,2), (3,4), (5,2q)
Solution
The\quad equation\quad of\quad a\quad line\quad with\quad two\quad given\quad points\quad is\quad given\quad as\quad \\ \frac { y-{ y }_{ 1 } }{ x-{ x }_{ 1 } } =\frac { { y }_{ 2 }-{ y }_{ 1 } }{ x_{ 2 }-{ x }_{ 1 } } \\ For\quad line\quad PQ,\quad \frac { y-q }{ x-1 } =\frac { 2-q }{ 3-1 } \\ 2(y-q)=(x-1)(2-q)\\ 2y-2q=2x-qx-2+q\\ 2y=(2-q)x-2+3q\\ y=\frac { 2-q }{ 2 } x+\frac { 3q }{ 2 } -1\\ { m }_{ 1 }=\frac { 2-q }{ 2 } \\ For\quad line\quad \quad RS,\quad \frac { y-4 }{ x-3 } =\frac { 2q-4 }{ 5-3 } \\ 2(y-4)=(x-3)(2q-4)\\ 2y-8=2qx-4x-6q+12\\ 2y=(2q-4)x-6q+20\\ y=(\frac { 2q-4 }{ 2 } )x-3q+10\\ y=(q-2)x-3q+10\\ { m }_{ 2 }=q-2\\ For\quad parallel\quad lines,{ m }_{ 1 }={ m }_{ 2 }\\ \frac { 2-q }{ 2 } =q-2\\ 2-q=2(q-2)\\ 2-q=2q-4\\ 3q=-6\\ q=-2
2007-10: If the lines 3y=4x-1 and 9y=x+3 are parallel to each other, the value of q is?
Solution
3y=4x-1\\ y=\frac { 4 }{ 3 } x-\frac { 1 }{ 3 } \\ { m }_{ 1 }=\frac { 4 }{ 3 } \\ 9y=x+3\\ y=\frac { x }{ q } +\frac { 3 }{ q } \\ { m }_{ 2 }=\frac { 1 }{ q } \\ For\quad parallelism,{ m }_{ 1 }={ m }_{ 2 }\\ \frac { 4 }{ 3 } =\frac { 1 }{ q } \\ q=\frac { 3 }{ 4 }
That’s the much we can take on this. If you have any question on this topic or related to mathematics, feel free to use the comment session. Be rest assured, there are mathematicians here. Also, notify us if you see any errors in the solutions above. Kindly share.
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