104 ANALYTIC GEOMETRY [Cii. V. 



9. Show that the triangle whose vertices are the points ('J, 1 ), ( 

 1) is a right triangle. 



10. Prove analytically that the perpendiculars envi.-d at tin- middle 

 points of the sides of tin- triangle, the equations of whose sides are 



x + y + 1 =0, 3z + 5y + 11 = 0, and x + 2y + 4 = 0, 

 meet in a point which is equidistant from the vertices. 



11. Kind the equations of the lines through the vertices and perj-n- 

 di.-ular to the opposite sides of the triangle in exercise 10. 1'rovr that 

 these lines also meet in a common point. 



12 A line passes through the point (2, -3) and is parallel to ill.- 

 HIM* through the two points (4, 7) and (-1, '.>) ; find its equation. 



13. Find the equation of the line which passes through the point of 

 intersection of the two lines 10* + 5y + ll=0, and x + 2y + 14 = 0, 

 and which is perpendicular to the line ar + 7y+l=0. 



This problem may be solved by first finding the point of intersection 

 (. - V) of the two given lines, and then, by formula [11] (see 

 Art. 62), writing the equation of the required line, viz. : 



which reduces to 7 x y = 31. 



The problem may also be solved somewhat more briefly, and much 

 more elegantly, by employing the theorem of Art. 41. By this theorem 

 the equation of the required line is of the form 



lOx + 5y + 11 + k(x + 2y + 14) = 0, 

 iX (10 + *)a?+(r> + 2t)y + ll + 14* = 0. 



It only remains to determine the constant , so that this line shall 

 be perpendicular tox + 7y+l=0. By Art. 62 its slope must be 



-J- = 7, hence - ^^ = 7, whence k = - 3. 



Substituting this value of k above, the required equation becomes 

 7 x y = 31, as before. 



14. By the second method of exercise 13 find the equation of the line 

 which passes through the point of intersection of the two lines 2z + y = 5 

 and x = 3y - 8, and which is: (1) parallel to thp HIM- \ >/ = *r + 1 ; 

 (2) perpendicular to this line ; (3) inclined at an angle of 60 to thia 



( 1) passes through the point (-], 3). 



15. Solve exercise 10 by the method of exercise 14. 



