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# 11 x1 t05 06 line through pt of intersection (2012)

## by Nigel Simmons, Teacher at Baulkham Hills High School on Mar 26, 2012

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## 11 x1 t05 06 line through pt of intersection (2012)Webinar Transcript

• Equation of a line through a point and intersection of another two lines
• Equation of a line through a point and intersection of another two linese.g. Find the equation of the line that passes through the intersection of 2x + y + 1= 0 and 3x + 5y – 9 = 0 and the point (1,2).
• Equation of a line through a point and intersection of another two linese.g. Find the equation of the line that passes through the intersection of 2x + y + 1= 0 and 3x + 5y – 9 = 0 and the point (1,2). 2x  y 1  0 3x  5 y  9  0
• Equation of a line through a point and intersection of another two linese.g. Find the equation of the line that passes through the intersection of 2x + y + 1= 0 and 3x + 5y – 9 = 0 and the point (1,2). 2x  y 1  0 10 x  5 y  5  3x  5 y  9  0 3x  5 y  9
• Equation of a line through a point and intersection of another two linese.g. Find the equation of the line that passes through the intersection of 2x + y + 1= 0 and 3x + 5y – 9 = 0 and the point (1,2). 2x  y 1  0 10 x  5 y  5 ()  3x  5 y  9  0 3x  5 y  9
• Equation of a line through a point and intersection of another two linese.g. Find the equation of the line that passes through the intersection of 2x + y + 1= 0 and 3x + 5y – 9 = 0 and the point (1,2). 2x  y 1  0 10 x  5 y  5 ()  3x  5 y  9  0 3x  5 y  9 7x =  14
• Equation of a line through a point and intersection of another two linese.g. Find the equation of the line that passes through the intersection of 2x + y + 1= 0 and 3x + 5y – 9 = 0 and the point (1,2). 2x  y 1  0 10 x  5 y  5 ()  3x  5 y  9  0 3x  5 y  9 7x =  14 x  2
• Equation of a line through a point and intersection of another two linese.g. Find the equation of the line that passes through the intersection of 2x + y + 1= 0 and 3x + 5y – 9 = 0 and the point (1,2). 2x  y 1  0 10 x  5 y  5 ()  3x  5 y  9  0 3x  5 y  9 7x =  14 x  2  2  2   y  1  0 y3
• Equation of a line through a point and intersection of another two linese.g. Find the equation of the line that passes through the intersection of 2x + y + 1= 0 and 3x + 5y – 9 = 0 and the point (1,2). 2x  y 1  0 10 x  5 y  5 ()  3x  5 y  9  0 3x  5 y  9 7x =  14 x  2  2  2   y  1  0 y3  the lines intersect at  2,3
• Equation of a line through a point and intersection of another two linese.g. Find the equation of the line that passes through the intersection of 2x + y + 1= 0 and 3x + 5y – 9 = 0 and the point (1,2). 2x  y 1  0 10 x  5 y  5 ()  3x  5 y  9  0 3x  5 y  9 7x =  14 x  2  2  2   y  1  0 y3 3 2  the lines intersect at  2,3 m 2  1 1  3
• Equation of a line through a point and intersection of another two linese.g. Find the equation of the line that passes through the intersection of 2x + y + 1= 0 and 3x + 5y – 9 = 0 and the point (1,2). 2x  y 1  0 10 x  5 y  5 ()  3x  5 y  9  0 3x  5 y  9 7x =  14 x  2  2  2   y  1  0 y3 3 2 1 y  2    x  1  the lines intersect at  2,3 m 2  1 3 1  3
• Equation of a line through a point and intersection of another two linese.g. Find the equation of the line that passes through the intersection of 2x + y + 1= 0 and 3x + 5y – 9 = 0 and the point (1,2). 2x  y 1  0 10 x  5 y  5 ()  3x  5 y  9  0 3x  5 y  9 7x =  14 x  2  2  2   y  1  0 y3 3 2 1 y  2    x  1  the lines intersect at  2,3 m 2  1 3 1 3y  6  x 1  3 x  3y  7  0
• Alternatively
• Alternatively a1 x  b1 y  c1  k  a2 x  b2 y  c2   0
• Alternatively a1 x  b1 y  c1  k  a2 x  b2 y  c2   0 2 x  y  1  k  3x  5 y  9   0
• Alternatively a1 x  b1 y  c1  k  a2 x  b2 y  c2   0 2 x  y  1  k  3x  5 y  9   0 1, 2  : 2 1   2   1  k  3 1  5  2   9   0
• Alternatively a1 x  b1 y  c1  k  a2 x  b2 y  c2   0 2 x  y  1  k  3x  5 y  9   0 1, 2  : 2 1   2   1  k  3 1  5  2   9   0 5  4k  0
• Alternatively a1 x  b1 y  c1  k  a2 x  b2 y  c2   0 2 x  y  1  k  3x  5 y  9   0 1, 2  : 2 1   2   1  k  3 1  5  2   9   0 5  4k  0 4k  5 5 k  4
• Alternatively a1 x  b1 y  c1  k  a2 x  b2 y  c2   0 2 x  y  1  k  3x  5 y  9   0 1, 2  : 2 1   2   1  k  3 1  5  2   9   0 5  4k  0 4k  5 5 k  4 5 2x  y 1  3x  5 y  9   0 4
• Alternatively a1 x  b1 y  c1  k  a2 x  b2 y  c2   0 2 x  y  1  k  3x  5 y  9   0 1, 2  : 2 1   2   1  k  3 1  5  2   9   0 5  4k  0 4k  5 5 k  4 5 2x  y 1  3x  5 y  9   0 48 x  4 y  4  15 x  25 y  45  0
• Alternatively a1 x  b1 y  c1  k  a2 x  b2 y  c2   0 2 x  y  1  k  3x  5 y  9   0 1, 2  : 2 1   2   1  k  3 1  5  2   9   0 5  4k  0 4k  5 5 k  4 5 2x  y 1  3x  5 y  9   0 48 x  4 y  4  15 x  25 y  45  0 7 x  21 y  49  0
• Alternatively a1 x  b1 y  c1  k  a2 x  b2 y  c2   0 2 x  y  1  k  3x  5 y  9   0 1, 2  : 2 1   2   1  k  3 1  5  2   9   0 5  4k  0 4k  5 5 k  4 5 2x  y 1  3x  5 y  9   0 48 x  4 y  4  15 x  25 y  45  0 7 x  21 y  49  0 x  3y  7  0
• Alternatively a1 x  b1 y  c1  k  a2 x  b2 y  c2   0 2 x  y  1  k  3x  5 y  9   0 1, 2  : 2 1   2   1  k  3 1  5  2   9   0 5  4k  0 4k  5 5 k  4 5 2x  y 1  3x  5 y  9   0 48 x  4 y  4  15 x  25 y  45  0 Exercise 5F; 2b, 3b, 6b(i), 7ab (i, iii), 9, 10, 13* 7 x  21 y  49  0 x  3y  7  0 Exercise 5G; 2 to 14 evens, 15*